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CBSE Class 12 Mathematics Chapter 7 Integrals – NCERT Solutions Exercise 7.5 [2025-26]

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Download Free PDF of Integrals Exercise 7.5 Solutions for Class 12 Maths

You’ve reached the most trusted resource for NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.5, designed specifically for the CBSE 2024-25 syllabus. Chapter 7, Integrals, contributes up to 9 marks on the board exam, making your conceptual clarity here essential for strong performance. These solutions guide you step by step through advanced integration techniques using partial fractions, substitution, and standard formula applications.

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If you’ve searched for “exercise 7.5 class 12” or “class 12 maths ex 7.5 solutions,” you’ll find every question solved in detail to strengthen your integration techniques and board exam readiness. With semantic keywords like “practice integration sums” and “stepwise integration method,” this guide aims to improve your calculation accuracy and build exam-day confidence.


Rely on Vedantu’s comprehensive, syllabus-aligned answers to master each integration type quickly. For advanced revision, you can also practice with the Class 12 Maths Chapter 7 Miscellaneous Exercise. Your focused effort on these solutions can help you achieve high marks in the final exam.

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Access NCERT Solutions for Class 12 Maths Chapter 7 – Integrals Exercise 7.5

Exercise 7.5

1. Integrate $\dfrac{x}{{(x + 1)(x + 2)}}$

Ans: Let $\dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$ \Rightarrow x = A(x + 2) + B(x + 1)$

Equating the coefficients of ${\text{x}}$ and constant term, we obtain $A + B = 1$

$2A + B = 0$

On solving. we obtain ${\text{A}} =  - 1$ and ${\text{B}} = 2$

$\therefore \dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{{ - 1}}{{(x + 1)}} + \dfrac{2}{{(x + 2)}}$

$ \Rightarrow \int {\dfrac{x}{{(x + 1)(x + 2)}}} dx = \int {\dfrac{{ - 1}}{{(x + 1)}}}  + \dfrac{2}{{(x + 2)}}dx$

$ =  - \log |x + 1| + 2\log |x + 2| + C$

$ = \log {(x + 2)^2} - \log |x + 1| + C$

$ = \log \dfrac{{{{(x + 2)}^2}}}{{(x + 1)}} + C$

Where $C$ is an arbitrary constant

2. Integrate $\dfrac{1}{{{x^2} - 9}}$

Ans: Let $\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{A}{{(x + 3)}} + \dfrac{B}{{(x - 3)}}$

$1 = A(x - 3) + B(x + 3)$

Equating the coefficients of $x$ and constant term, we obtain $A + B = 0$

$ - 3A + 3B = 1$

On solving. we obtain

${\text{A}} =  - \dfrac{1}{6}$ and ${\text{B}} = \dfrac{1}{6}$

$\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}$

$ \Rightarrow \int {\dfrac{1}{{\left( {{x^2} - 9} \right)}}} dx = \int {\left( {\dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}} \right)} dx$

$ =  - \dfrac{1}{6}\log |x + 3| + \dfrac{1}{6}\log |x - 3| + C$

$ = \dfrac{1}{6}\log \dfrac{{|(x - 3)|}}{{|(x + 3)|}} + C$

Where $C$ is an arbitrary constant

3. Integrate $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}$

Ans: Let $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$3x - 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$ - 5A - 4B - 3C = 3$

$6A + 3B + 2C =  - 1$

Solving these equations, we obtain ${\text{A}} = 1,\;{\text{B}} =  - 5$, and $C = 4$

$\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}$

$ \Rightarrow \int {\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}} \right\}} dx$

$ = \log |x - 1| - 5\log |x - 2| + 4\log |x - 3| + C$

Where $C$ is an arbitrary constant.

4. Integrate $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}$

Ans: $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$45 - 3C = 1$

$6A + 4B + 2C = 0$

Solving these equations, we obtain $A = \dfrac{1}{2},B = 2$ and $C = \dfrac{3}{2}$

$\therefore \dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}$

$ \Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}} \right\}} dx$

$ = \dfrac{1}{2}\log |x - 1| - 2\log |x - 2| + \dfrac{3}{2}\log |x - 3| + C$

Where $C$ is an arbitrary constant.

5. Integrate $\dfrac{{2x}}{{{x^2} + 3x + 2}}$

Ans: Let $\dfrac{{2x}}{{{x^2} + 3x + 2}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$2x = A(x + 2) + B(x + 1)$

$ \ldots (1)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B = 2$

$2A + B = 0$

Solving these equations, we obtain $A =  - 2$ and ${\mathbf{B}} = 4$

$\therefore \dfrac{{2x}}{{(x + 1)(x + 2)}} = \dfrac{{ - 2}}{{(x + 1)}} + \dfrac{4}{{(x + 2)}}$

$ \Rightarrow \int {\dfrac{{2x}}{{(x + 1)(x + 2)}}} dx = \int {\left\{ {\dfrac{4}{{(x + 2)}} - \dfrac{2}{{(x + 1)}}} \right\}} dx$

$ = 4\log |x + 2| - 2\log |x + 1| + C$

Where $C$ is an arbitrary constant.

6. Integrate $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {1 - {x^2}} \right)$ by $x(1 - 2x)$, we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{{2 - x}}{{x(1 - 2x)}}} \right)$

Let $\dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{A}{x} + \dfrac{B}{{(1 - 2x)}}$

$ \Rightarrow (2 - x) = A(1 - 2x) + Bx$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $ - 2A + B =  - 1$

And $A = 2$ Solving these equations, we obtain $A = 2$ and $B = 3$ $\therefore \dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{2}{x} + \dfrac{3}{{1 - 2x}}$

Substituting in equation (1), we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left\{ {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right\}$

$\int {\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}} dx = \int {\left\{ {\dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right)} \right\}} dx$

$ = \dfrac{x}{2} + \log |x| + \dfrac{3}{{2( - 2)}}\log |1 - 2x| + C$

$ = \dfrac{x}{2} + \log |x| - \dfrac{3}{4}\log |1 - 2x| + c$

Where $C$ is an arbitrary constant.

7. Integrate $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}$

Ans: Let $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \dfrac{C}{{(x - 1)}}$

$x = (Ax + B)(x - 1) + C\left( {{x^2} + 1} \right)$

$x = A{x^2} - Ax + Bx - B + C{x^2} + C$

Equating the coefficients of ${x^2},x$, and constant term, we obtain

A $ - A + B = 1$

$ - B + C = 0$

On solving these equations, we obtain ${\text{A}} =  - \dfrac{1}{2},\;{\text{B}} = \dfrac{1}{2}$, and ${\text{C}} = \dfrac{1}{2}$

From equation (1), vre obtain $\therefore \dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{\left( { - \dfrac{1}{2}x + \dfrac{1}{2}} \right)}}{{{x^2} + 1}} + \dfrac{{\dfrac{1}{2}}}{{(x - 1)}}$

$ \Rightarrow \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}}  =  - \dfrac{1}{2}\int {\dfrac{x}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx$

$ =  - \dfrac{1}{4}\int {\dfrac{{2x}}{{{x^2} + 1}}} dx + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

Consider $\int {\dfrac{{2x}}{{{x^2} + 1}}} dx$, let $\left( {{x^2} + 1} \right) = t \Rightarrow 2xdx = dt$

$ \Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{{dt}}{t}}  - \log |t| - \log \left| {{x^2} + 1} \right|$

$\therefore \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}}  =  - \dfrac{1}{4}\log \left| {{x^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

$ = \dfrac{1}{2}\log |{\text{x}} - 1| - \dfrac{1}{4}\log \left| {{{\text{x}}^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}{\text{x}} + {\text{C}}$

Where $C$ is an arbitrary constant.

8. Integrate $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}$

Ans: Let $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}} - \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 2)}}$

$x = A(x - 1)(x + 2) + B(x + 2) + C{(x - 1)^2}$

Equating the coefficients of ${{\text{x}}^2},{\text{x}}$ and constant term, we obtain ${\text{A}} + {\text{C}} = 0$

$A + B - 2C = 1$

$ - 2\;{\text{A}} + 2\;{\text{B}} + {\text{C}} = 0$

On solving. we obtain $A = \dfrac{2}{9}$ and $C = \dfrac{{ - 2}}{9}$

${\text{B}} = \dfrac{1}{3}$

$\therefore \dfrac{x}{{{{(x - 1)}^2}(x + 2)}} = \dfrac{2}{{9(x - 1)}} + \dfrac{1}{{3{{(x - 1)}^2}}} - \dfrac{2}{{9(x + 2)}}$

$ \Rightarrow \int {\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x - 1)}}} dx + \dfrac{1}{3}\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x + 2)}}} dx$

$ = \dfrac{2}{9}\log |x - 1| + \dfrac{1}{3}\left( {\dfrac{{ - 1}}{{x - 1}}} \right) - \dfrac{2}{9}\log |x + 2| + C$

$ = \dfrac{2}{9}\log \left| {\dfrac{{x - 1}}{{x + 2}}} \right| - \dfrac{1}{{3(x - 1)}} + C$

Where $C$ is an arbitrary constant.

9. Integrate $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}$

Ans: $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}} = \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}$

Let $\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 1)}}$

$3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C{(x - 1)^2}$

$3x + 5 = A\left( {{x^2} - 1} \right) + B(x + 1) + C\left( {{x^2} + 1 - 2x} \right)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + C = 0$

$B - 2C - 3$

$ - A + B + C = 5$

On solving. we obtain $B = 4$

${\text{A}} =  - \dfrac{1}{2}$ and ${\text{C}} = \dfrac{1}{2}$

$\therefore \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{{ - 1}}{{2(x - 1)}} + \dfrac{4}{{{{(x - 1)}^2}}} + \dfrac{1}{{2(x + 1)}}$

$ \Rightarrow \int {\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}} dx =  - \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx + 4\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx$

$ =  - \dfrac{1}{2}\log |x - 1| + 4\left( {\dfrac{{ - 1}}{{x - 1}}} \right) + \dfrac{1}{2}\log |x + 1| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{x + 1}}{{x - 1}}} \right| - \dfrac{4}{{(x - 1)}} + C$

Where $C$ is an arbitrary constant.

10. Integrate $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}$

Ans: $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}} = \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}}$

Let $\dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{C}{{(2x + 3)}}$

$ \Rightarrow (2x - 3) = A(x - 1)(2x + 3) + B(x + 1)(2x + 3) + C(x + 1)(x - 1)$

$ \Rightarrow (2x - 3) = A\left( {2{x^2} + x - 3} \right) + B\left( {2{x^2} + 5x + 3} \right) + C\left( {{x^2} - 1} \right)$

$ \Rightarrow (2x - 3) = (2A + 2B + C){x^2} + (A + 5B)x + ( - 3A + 3B - C)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $2A + 2B + C = 0$

$A + 5B = 2$

$ - 3A + 3B - C =  - 3$

On solving, we obtain ${\text{B}} =  - \dfrac{1}{{10}},\;{\text{A}} = \dfrac{5}{2}$, and ${\text{C}} =  - \dfrac{{24}}{5}$

$\therefore \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{5}{{2(x + 1)}} - \dfrac{1}{{10(x - 1)}} - \dfrac{{24}}{{5(2x + 3)}}$

$ \Rightarrow \int {\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}} dx = \dfrac{5}{2}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{1}{{10}}\int {\dfrac{1}{{x - 1}}} dx - \dfrac{{24}}{5}\int {\dfrac{1}{{(2x + 3)}}} dx$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{24}}{{5 \times 2}}\log |2x + 3|$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{12}}{5}\log |2x + 3| + C$

Where $C$ is an arbitrary constant.

11. Integrate $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}$

Ans : $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}} = \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}}$

Let $\dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}} + \dfrac{C}{{(x - 2)}}$

$5x = A(x + 2)(x - 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $A + B + C = 0$

$B + 3C = 5$ and $4A - 2B + 2C = 0$

On solving. we obtain $A - \dfrac{5}{3},B -  - \dfrac{5}{2}$, and $C - \dfrac{5}{6}$

$\therefore \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{5}{{3(x + 1)}} +  - \dfrac{5}{{2(x + 2)}} + \dfrac{5}{{6(x - 2)}}$

$ \Rightarrow \int {\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}} dx = \dfrac{5}{3}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{5}{2}\int {\dfrac{1}{{(x + 2)}}} dx + \dfrac{5}{6}\int {\dfrac{1}{{(x - 2)}}} dx$

$ = \dfrac{5}{3}\log |x + 1| - \dfrac{5}{2}\log |x + 2| + \dfrac{5}{6}\log |x - 2| + C$

Where $C$ is an arbitrary constant.

12. Integrate $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {{x^3} + x + 1} \right)$ by ${x^2} - 1$, we obtain $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{{2x + 1}}{{{x^2} - 1}}$

Let $\dfrac{{2x + 1}}{{{x^2} - 1}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}}$

$2x + 1 = A(x - 1) + B(x + 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 2$

$ - A + B = 1$

On solving. we obtain $A - \dfrac{1}{2}$ and $B - \dfrac{3}{2}$

$\therefore \dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{1}{{2(x + 1)}} + \dfrac{3}{{2(x - 1)}}$

$ \Rightarrow \int {\dfrac{{{x^3} + x + 1}}{{{x^2} + 1}}} dx = \int x dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx + \dfrac{3}{2}\int {\dfrac{1}{{(x - 1)}}} dx$

$ = \dfrac{{{x^2}}}{2} + \log |x + 1| + \dfrac{3}{2}\log |x - 1| + C$

Where $C$ is an arbitrary constant.

13. Integrate $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}$

Ans:

Let $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(1 - x)}} + \dfrac{{Bx + C}}{{\left( {1 + {x^2}} \right)}}$

$2 = A\left( {1 + {x^2}} \right) + (Bx + C)(1 - x)$

$2 = A + A{x^2} + Bx - B{x^2} + C - Cx$

Equating the coefficient of ${x^2},x$, and constant term, we obtain ${\text{A}} - {\text{B}} = 0$

${\mathbf{B}} - {\mathbf{C}} = {\mathbf{0}}$

$A + C = 2$

On solving these equations, we obtain $A = 1,B = 1$, and $C = 1$

$\therefore \dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{1}{{1 - x}} + \dfrac{{x + 1}}{{1 + {x^2}}}$

$ \Rightarrow \int {\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}} dx = \int {\dfrac{1}{{1 - x}}} dx + \int {\dfrac{x}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \int {\dfrac{1}{{1 - x}}} dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \log |x - 1| + \dfrac{1}{2}\log \left| {1 + {x^2}} \right| + {\tan ^{ - 1}}x + C$

Where $C$ is an arbitrary constant.

14. Integrate $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}$

Ans:

Let $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{{{(x + 2)}^2}}}$

$ \Rightarrow 3x - 1 = A(x + 2) + B$

Equating the coefficient of $x$ and constant term, we obtain $A = 3$

$2A + B =  - 1 \Rightarrow \mathbb{B} -  - 7$

$\therefore \dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{3}{{(x + 2)}} - \dfrac{7}{{{{(x + 2)}^2}}}$

$ \Rightarrow \int {\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}} dx = 3\int {\dfrac{1}{{(x + 2)}}} dx - 7\int {\dfrac{1}{{{{(x + 2)}^2}}}} dx$

$ = 3\log |x + 2| - 7\left( {\dfrac{{ - 1}}{{(x + 2)}}} \right) + C$

$ = 3\log |x + 2| + \dfrac{7}{{(x + 2)}} + C$

Where $C$ is an arbitrary constant.

15. Integrate $\dfrac{1}{{{x^4} - 1}}$

Ans :

$\dfrac{1}{{\left( {{x^4} - 1} \right)}} - \dfrac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} - \dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}}$

Let $\dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 1} \right)}}$

$1 = A(x - 1)\left( {1 + {x^2}} \right) + B(x + 1)\left( {1 + {x^2}} \right) + (Cx + D)\left( {{x^2} - 1} \right)$

$1 = A\left( {{x^3} + x - {x^2} - 1} \right) + B\left( {{x^3} + x + {x^2} + 1} \right) + C{x^3} + D{x^2} - Cx - D$

$1 = (A + B + C){x^3} + ( - A + B + D){x^2} + (A + B - C)x + ( - A + B - D)$

Equating the coefficient of ${x^3},{x^2},x$, and constant term, we obtain $A + B + C = 0$

$ - A + B + D = 0$

$A + B - C = 0$

$ - A + B - D = 1$

${\text{A}} =  - \dfrac{1}{4},\;{\text{B}} = \dfrac{1}{4},{\text{C}} = {\text{O}}$, and ${\text{D}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{\left( {{x^n} - 1} \right)}} = \dfrac{{ - 1}}{{4(x + 1)}} + \dfrac{1}{{4(x - 1)}} + \dfrac{1}{{2\left( {{x^2} + 1} \right)}}$

$ \Rightarrow \int {\dfrac{1}{{{x^4} - 1}}} dx -  - \dfrac{1}{4}\log |x - 1| + \dfrac{1}{4}\log |x - 1| - \dfrac{1}{2}{\tan ^1}x + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| - \dfrac{1}{2}{\tan ^1}x + C$

Where $C$ is an arbitrary constant.

16. Integrate $\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Hint: multiply numerator and denominator by ${x^{n - 1}}$ and put $x^n = t$

Ans:

 $\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Multiplying numerator and denominator by ${x^{n - 1}}$, we obtain $\dfrac{1}{{x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^{n - 1}}x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}$

Let ${x^n} = t \Rightarrow n{x^{ - 1}}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}} dx = \int {\dfrac{{{x^{n - 1}}}}{{{x^7}\left( {{x^n} + 1} \right)}}} dx = \dfrac{1}{n}\int {\dfrac{1}{{t(t + 1)}}} dt$

Let $\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t + 1)}}$

$1 = A(1 + t) + Bt$

Equating the coefficients of $t$ and constant, ve obtain $A = 1$ and $B =  - 1$

$\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{(1 + t)}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} + 1} \right)}}} dx = \dfrac{1}{n}\int {\left\{ {\dfrac{1}{t} - \dfrac{1}{{(1 + t)}}} \right\}} dx$

$\dfrac{1}{n}[\log |t| - \log |t + 1|] + C$

$ = \dfrac{1}{n}\left[ {\log \left| {{x^n}} \right| - \log \left| {{x^n} + 1} \right|} \right] + C$

$ = \dfrac{1}{n}\log \left| {\dfrac{{{x^n}}}{{{x^2} + 1}}} \right| + C$

Where $C$ is an arbitrary constant.

17. Integrate $\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

(Hint: Put $\sin x = t]$

Ans:

$\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

let $\sin x = t \Rightarrow \cos xdx = dt$

$\therefore \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} $

Let $\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}$

$1 = A(2 - t) + B(1 - t)$

Equating the coefficients of $t$ and constant, wre obtain $ - A - B = 0$ and $2A + B = 1$

On solving. we obtain $A = 1$ and $B =  - 1$

$\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}$

$ \Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\left\{ {\dfrac{1}{{1 - t}} - \dfrac{1}{{(2 - t)}}} \right\}} dt$

$ = \log \left| {\dfrac{{2 - t}}{{1 - t}}} \right| + C$

$ = \log \left| {\dfrac{{2 - \sin x}}{{1 - \sin x}}} \right| + C$

Where $C$ is an arbitrary constant.

18. Integrate $\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Ans:

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \dfrac{{4{x^2} + 10}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Let $\dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 3} \right)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}$

$4{x^2} + 10 = (Ax + B)\left( {{x^2} + 4} \right) + (Cx + D)\left( {{x^2} + 3} \right)$

$4{x^2} + 10 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + 3Cx + D{x^2} + 3D$

$4{x^2} + 10 = (A + C){x^2} + (B + D){x^2} + (4A + 3C)x + (4B + 3D)$

Equating the coefficients of ${x^3},{x^2},x$ and constant term, we obtain $A + C = 0$

$B + D = 4$

$4A + 3C = 0$

$4B + 3D = 10$

On solving these equations, we obtain ${\text{A}} = 0,\;{\text{B}} =  - 2,{\text{C}} = 0$, and ${\text{D}} = 6$

$\therefore \dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}$

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \left( {\dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right)$

$ \Rightarrow \int {\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}} dx = \int {\left\{ {1 + \dfrac{2}{{\left( {{x^2} + 3} \right)}} - \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right\}} dx$

$ = \int {\left\{ {1 + \dfrac{2}{{{x^2} + {{(\sqrt 3 )}^2}}} - \dfrac{6}{{{x^2} + {2^2}}}} \right\}} $

$ = x + 2\left( {\dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{x}{{\sqrt 3 }}} \right) - 6\left( {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{x}{2}} \right) + C$

$ = x + \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\dfrac{x}{2} + C$

Where $C$ is an arbitrary constant.

19. Integrate $\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Ans:

$\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Let ${x^2} = t \Rightarrow 2xdx = dt$

$\therefore \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\dfrac{{dt}}{{(t + 1)(t + 3)}}} $

Let $\dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{A}{{(t + 1)}} + \dfrac{B}{{(t + 3)}}$

$1 = A(t + 3) + B(t + 1)$

Equating the coefficients of ${\text{t}}$ and constant, we obtain $A + B = 0$ and $3A + B = 1$

On solving. we obtain ${\text{A}} = \dfrac{1}{2}$ and ${\text{B}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}$

$ \Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\left\{ {\dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}} \right\}} dt$

$ = \dfrac{1}{2}\log |(t + 1)| - \dfrac{1}{2}\log |t + 3| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{t + 1}}{{t + 3}}} \right| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C$

Where $C$ is an arbitrary constant.

20. Integrate $\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Ans:

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Multiplying numerator and denominator by ${x^3}$, we obtain

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}} = \dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \int {\dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}} dx$

Let ${x^4} = t \Rightarrow 4{x^3}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\dfrac{{dt}}{{t(t - 1)}}} $

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t - 1)}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of $t$ and constant, we obtain $A + B = 0$ and $ - A = 1$

${\text{A}} =  - 1$ and ${\mathbf{B}} = 1$

$ \Rightarrow \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\left\{ {\dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}} \right\}} dt$

$ - \dfrac{1}{4}[ - \log |t| + \log |t - 1|] + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{{x^4} - 1\mid }}{{{x^4}}}} \right| + C$

Where $C$ is an arbitrary constant.

21. Integrate $\dfrac{1}{{\left( {{{\text{e}}^ x } - 1} \right)}}$

Hint: Put $\dfrac{1}{e^x - 1}$

Ans :

Let ${e^x} = t \Rightarrow {e^x}dx = dt$

$ \Rightarrow \int {\dfrac{1}{{\left( {{{\text{e}}^x} - 1} \right)}}} {\text{dx}} = \int {\dfrac{1}{{{\text{t}} - 1}}}  \times \dfrac{{{\text{dt}}}}{{\text{t}}} - \int {\dfrac{1}{{{\text{t}}({\text{t}} - 1)}}} {\text{dt}}$

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{t - 1}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of ${\text{t}}$ and constant, we obtain

$A + B = 0$ and $ - A = 1$

$A =  - 1$ and ${\mathbf{B}} = {\mathbf{1}}$

$\therefore \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$ \Rightarrow \int {\dfrac{1}{{t(t - 1)}}} dt = \log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$ = \log \left| {\dfrac{{{{\text{e}}^x} - 1}}{{{{\text{e}}^x}}}} \right| + {\text{C}}$

Where $C$ is an arbitrary constant.

22. $\int {\dfrac{{xdx}}{{(x - 1)(x - 2)}}} $ equals

a. $A\log \left| {\dfrac{{{{(x - 1)}^2}}}{{x - 2}}} \right| + C$

b. $\log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

c. $\log \left| {{{\left( {\dfrac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$

d. $\log |(x - 1)(x - 2)| + C$

Ans:

Let $\dfrac{x}{{(x - 1)(x - 2)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}}$

$x = A(x - 2) + B(x - 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 1$ and $ - 2A - B = 0$

$A--1$ and $B = 2$

$\dfrac{x}{{(x - 1)(x - 2)}} =  - \dfrac{1}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}$

$ \Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)}}} dx = \int {\left\{ {\dfrac{{ - 1}}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}} \right\}} dx$

$ =  - \log |x - 1| + 2\log |x - 2| + C$

$ = \log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

Hence, the correct Answer is $B$.

23. $\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} $ equals

a. $\log |x| - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

b. $\log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

c. $ - \log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

d. $\dfrac{1}{2}\log |x| + \log \left( {{x^2} + 1} \right) + C$

Ans :

Let $\dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$

$1 = A\left( {{x^2} + 1} \right) + (Bx + C)x$

Equating the coefficients of ${x^2},x$, and constant term, we obtain $A + B = 0$

$c = 0$

$A = 1$

On solving these equations, we obtain $A = 1,B =  - 1$, and $C = 0$

$\therefore \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}}} \right\}} dx$

$ - \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Hence, the correct Answer is ${\text{A}}$. 

Where $C$ is an arbitrary constant.

Alter:

$ \Rightarrow \int {\dfrac{1}{{{\text{x}}\left( {{{\text{x}}^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{{\text{x}}}{{{{\text{x}}^2}\left( {{{\text{x}}^2} + 1} \right)}}} \right\}} {\text{dx}}$

Let ${x^2} = t$, therefore, $2xdx = dt$ $\therefore \int {\dfrac{x}{{{x^2}\left( {{x^2} + 1} \right)}}} dx = \dfrac{1}{2}\int {\dfrac{{dt}}{{t(t + 1)}}}  = \dfrac{1}{2}\int {\dfrac{{(t + 1) - t}}{{t(t + 1)}}} dt = \dfrac{1}{2}\int \frac{1}{t} -  \dfrac{1}{{t + 1}}dt$

$ - \dfrac{1}{2}[\log t - \log (t + 1)] + C$

$ = \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Where $C$ is an arbitrary constant.


Conclusion

The NCERT Solutions for Class 12 Chapter 7 Integrals Exercise 7.5 focuses on the application of integration techniques to solve various types of integral problems, particularly focusing on integration by parts and the integration of rational functions. It's essential to understand the fundamental integration formulas and the steps for applying integration by parts, as these are commonly used methods in this exercise. Emphasizing practice on these techniques will help in solving the problems efficiently.


Class 12 Maths Chapter 7: Exercises Breakdown

S.No.

Chapter 7 - Integrals Exercises in PDF Format

1

Class 12 Maths Chapter 7 Exercise 7.1 - 22 Questions & Solutions (21 Short Answers, 1 MCQs)

2

Class 12 Maths Chapter 7 Exercise 7.2 - 39 Questions & Solutions (37 Short Answers, 2 MCQs)

3

Class 12 Maths Chapter 7 Exercise 7.3 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

4

Class 12 Maths Chapter 7 Exercise 7.4 - 25 Questions & Solutions (23 Short Answers, 2 MCQs)

5

Class 12 Maths Chapter 7 Exercise 7.6 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

6

Class 12 Maths Chapter 7 Exercise 7.7 - 11 Questions & Solutions (9 Short Answers, 2 MCQs)

7

Class 12 Maths Chapter 7 Exercise 7.8 - 6 Questions & Solutions (6 Short Answers)

8

Class 12 Maths Chapter 7 Exercise 7.9 - 22 Questions & Solutions (20 Short Answers, 2 MCQs)

9

Class 12 Maths Chapter 7 Exercise 7.10 - 10 Questions & Solutions (8 Short Answers, 2 MCQs)

10

Class 12 Maths Chapter 7 Miscellaneous Exercise - 40 Questions & Solutions



CBSE Class 12 Maths Chapter 7 Other Study Materials



NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.




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FAQs on CBSE Class 12 Mathematics Chapter 7 Integrals – NCERT Solutions Exercise 7.5 [2025-26]

1. What is the primary integration technique taught in the NCERT Solutions for Class 12 Maths, Chapter 7, Exercise 7.5?

The primary technique covered in Exercise 7.5 for the 2025-26 syllabus is integration by partial fractions. This method is used to integrate rational functions, which are expressions in the form of a ratio of two polynomials P(x)/Q(x).

2. How can a student identify when to use the partial fractions method for solving an integral?

You should use the partial fractions method when the integrand is a rational function where the denominator can be factorised into linear or quadratic factors. If the expression cannot be integrated directly or simplified easily using substitution, partial fractions is the correct approach prescribed in the NCERT solutions.

3. What is the crucial first step in the NCERT method if the integrand is an improper rational function?

If the degree of the numerator polynomial is greater than or equal to the degree of the denominator, it is an improper fraction. The crucial first step is to perform polynomial long division. This converts the integrand into the sum of a polynomial and a proper rational function, which can then be integrated using standard formulas and the partial fraction method.

4. How are partial fractions set up for different denominator forms as per the CBSE 2025-26 syllabus?

The NCERT solutions for Chapter 7 detail the setup based on the factors in the denominator:

  • Non-repeated linear factors: For a term like (x-a)(x-b), the setup is A/(x-a) + B/(x-b).
  • Repeated linear factors: For a term like (x-a)², the setup is A/(x-a) + B/(x-a)².
  • Irreducible quadratic factors: For a term like (x-a)(x²+bx+c), the setup is A/(x-a) + (Bx+C)/(x²+bx+c).

5. What are the step-by-step methods to find the values of constants (A, B, C) in a partial fraction decomposition?

After setting up the partial fraction equation, you can find the constants by:

  • Equating Coefficients: Multiply out the terms, group them by powers of x (like x², x, and the constant term), and equate the coefficients on both sides of the equation to form a system of linear equations.
  • Substituting Values: Substitute strategic values of x (especially the roots of the denominator) to make certain terms zero, which simplifies the process of finding the constants directly.

6. After splitting an expression into partial fractions, what standard formulas are used to get the final answer?

Once the integrand is decomposed, the NCERT solutions rely on these standard integration formulas:

  • ∫(1/(ax+b)) dx = (1/a) log|ax+b| + C
  • ∫(1/(x-a)²) dx = -1/(x-a) + C
  • ∫(1/(x² + a²)) dx = (1/a) tan⁻¹(x/a) + C
  • ∫(x/(x² + a²)) dx = (1/2) log|x² + a²| + C

7. What are the most common mistakes students make when solving integrals using partial fractions?

Common errors to avoid when following the NCERT solutions for Chapter 7 include:

  • Forgetting to perform long division for improper rational functions first.
  • Using an incorrect form for the partial fraction decomposition (e.g., using A instead of Ax+B for a quadratic factor).
  • Making algebraic mistakes while solving for the constants A, B, and C.
  • Forgetting to add the constant of integration, C, in the final answer for indefinite integrals.

8. Why is the partial fraction method so effective for integrating rational functions?

The method is effective because it solves a fundamental problem: most complex rational functions do not have a direct integration formula. Partial fractions decompose a single, complicated fraction into a sum of several simpler fractions. Each of these simpler fractions can be easily integrated using basic, known formulas, thus making the overall problem solvable.

9. How does the solution process differ when applying partial fractions to a definite integral versus an indefinite integral?

The decomposition process is identical for both. The only difference is in the final step. For an indefinite integral, you integrate and add the constant 'C'. For a definite integral, after finding the antiderivative using partial fractions, you apply the Fundamental Theorem of Calculus by evaluating it at the upper and lower limits and subtracting the results, which gives a specific numerical answer.