Class 12 Maths NCERT Solutions Chapter 7 Ex 7.5 – Integration by Partial Fractions | Boundless Maths

Ex 7.5 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.5 – Integration by Partial Fractions

Chapter 7 Ex 7.5 covers integration by partial fractions — all 23 questions solved step by step, resolving rational functions into distinct linear, repeated linear, and irreducible quadratic pieces before integrating each one.

Every question here follows the same core routine: factor the denominator, write the fraction as a sum of simpler pieces with unknown constants, then solve for those constants by substituting convenient values of x or comparing coefficients. The exercise builds in difficulty by denominator type — distinct linear factors first (Q1–5), then repeated linear factors and irreducible quadratics (Q6–17), and finally denominators needing a substitution before decomposition, like eˣ − 1 or xⁿ + 1 (Q16, Q21). If the degree of the numerator is greater than or equal to the denominator, remember to divide first — skipping this step is the most common way this exercise catches students out. The two closing MCQs test the same decomposition skill under exam time pressure.

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23Questions

Medium–HardDifficulty Mix

2026-27CBSE Syllabus

Chapter 7 Ex 7.5 Solutions — All 23 Questions

Integrate: \dfrac{x}{(x+1)(x+2)}

Easy +

Solution

Split into partial fractions: \dfrac{x}{(x+1)(x+2)} = \dfrac{A}{x+1}+\dfrac{B}{x+2}, so x=A(x+2)+B(x+1).

Putting x=-1 gives A=-1; putting x=-2 gives B=2.

\displaystyle\int \dfrac{x}{(x+1)(x+2)}\,dx = 2\log|x+2| - \log|x+1| + C

Integrate: \dfrac{1}{x^2-9}

Easy +

Solution

Direct application of the standard formula \displaystyle\int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| with a=3.

\displaystyle\int \dfrac{1}{x^2-9}\,dx = \dfrac16\log\left|\dfrac{x-3}{x+3}\right| + C

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

Medium +

Solution

Write \dfrac{3x-1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}, so 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).

Putting x=1,2,3 in turn gives A=1, B=-5, C=4.

\displaystyle\int \dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx = \log|x-1| - 5\log|x-2| + 4\log|x-3| + C

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

Medium +

Solution

Same decomposition pattern as Q3: x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2). Putting x=1,2,3 gives A=\dfrac12, B=-2, C=\dfrac32.

\displaystyle\int \dfrac{x}{(x-1)(x-2)(x-3)}\,dx = \dfrac12\log|x-1| - 2\log|x-2| + \dfrac32\log|x-3| + C

Integrate: \dfrac{2x}{x^2+3x+2}

Easy +

Solution

Factor the denominator as (x+1)(x+2), then 2x=A(x+2)+B(x+1) gives A=-2 (at x=-1) and B=4 (at x=-2).

\displaystyle\int \dfrac{2x}{x^2+3x+2}\,dx = 4\log|x+2| - 2\log|x+1| + C

Integrate: \dfrac{1-x^2}{x(1-2x)}

Hard +

Solution

The numerator and denominator both have degree 2, so this is improper — divide first: \dfrac{1-x^2}{x-2x^2} = \dfrac12 + \dfrac{1-\frac{x}{2}}{x(1-2x)}, call the remainder term (1).

Decomposing (1) as \dfrac{A}{x}+\dfrac{B}{1-2x} and putting x=0 and x=\dfrac12 gives A=1, B=\dfrac32.

\displaystyle\int \dfrac{1-x^2}{x(1-2x)}\,dx = \dfrac{x}{2} + \log|x| - \dfrac34\log|1-2x| + C

Integrate: \dfrac{x}{(x^2+1)(x-1)}

Medium +

Solution

Write \dfrac{x}{(x^2+1)(x-1)} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}, so x=A(x^2+1)+(Bx+C)(x-1).

Putting x=1 gives A=\dfrac12; comparing coefficients gives B=-\dfrac12 and C=\dfrac12.

\displaystyle\int \dfrac{x}{(x^2+1)(x-1)}\,dx = \dfrac12\log|x-1| - \dfrac14\log(x^2+1) + \dfrac12\tan^{-1}x + C

Integrate: \dfrac{x}{(x-1)^2(x+2)}

Medium +

Solution

Write \dfrac{x}{(x-1)^2(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}, so x=A(x-1)(x+2)+B(x+2)+C(x-1)^2.

Putting x=1 gives B=\dfrac13; putting x=-2 gives C=-\dfrac29; comparing the x^2 coefficient gives A=\dfrac29.

\displaystyle\int \dfrac{x}{(x-1)^2(x+2)}\,dx = \dfrac29\log|x-1| - \dfrac{1}{3(x-1)} - \dfrac29\log|x+2| + C

Integrate: \dfrac{3x+5}{x^3-x^2-x+1}

Medium +

Solution

Factor by grouping: x^3-x^2-x+1 = x^2(x-1)-(x-1) = (x-1)^2(x+1).

Write \dfrac{3x+5}{(x-1)^2(x+1)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}, so 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2.

Putting x=1 gives B=4; putting x=-1 gives C=\dfrac12; comparing the x^2 coefficient gives A=-\dfrac12.

\displaystyle\int \dfrac{3x+5}{x^3-x^2-x+1}\,dx = \dfrac12\log|x+1| - \dfrac12\log|x-1| - \dfrac{4}{x-1} + C

Integrate: \dfrac{2x-3}{(x^2-1)(2x+3)}

Hard +

Solution

Factor x^2-1=(x-1)(x+1), then write \dfrac{2x-3}{(x-1)(x+1)(2x+3)} = \dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{C}{2x+3}, so 2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1).

Putting x=1 gives A=-\dfrac{1}{10}; putting x=-1 gives B=\dfrac52; putting x=-\dfrac32 gives C=-\dfrac{24}{5}.

\displaystyle\int \dfrac{2x-3}{(x^2-1)(2x+3)}\,dx = \dfrac52\log|x+1| - \dfrac{1}{10}\log|x-1| - \dfrac{12}{5}\log|2x+3| + C

Integrate: \dfrac{5x}{(x+1)(x^2-4)}

Medium +

Solution

Factor x^2-4=(x-2)(x+2), then 5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2). Putting x=-1,2,-2 gives A=\dfrac53, B=\dfrac56, C=-\dfrac52.

\displaystyle\int \dfrac{5x}{(x+1)(x^2-4)}\,dx = \dfrac53\log|x+1| + \dfrac56\log|x-2| - \dfrac52\log|x+2| + C

Integrate: \dfrac{x^3+x+1}{x^2-1}

Medium +

Solution

Improper — divide: x^3+x+1 = x(x^2-1)+(2x+1), so \dfrac{x^3+x+1}{x^2-1} = x+\dfrac{2x+1}{(x-1)(x+1)}.

Decomposing the remainder, putting x=1 gives A=\dfrac32; putting x=-1 gives B=\dfrac12.

\displaystyle\int \dfrac{x^3+x+1}{x^2-1}\,dx = \dfrac{x^2}{2} + \dfrac32\log|x-1| + \dfrac12\log|x+1| + C

Integrate: \dfrac{2}{(1-x)(1+x^2)}

Medium +

Solution

Write \dfrac{2}{(1-x)(1+x^2)} = \dfrac{A}{1-x}+\dfrac{Bx+C}{1+x^2}, so 2=A(1+x^2)+(Bx+C)(1-x).

Putting x=1 gives A=1; comparing coefficients gives B=1 and C=1.

\displaystyle\int \dfrac{2}{(1-x)(1+x^2)}\,dx = -\log|1-x| + \dfrac12\log(1+x^2) + \tan^{-1}x + C

Integrate: \dfrac{3x-1}{(x+2)^2}

Easy +

Solution

Write 3x-1=A(x+2)+B: comparing the x coefficient gives A=3, and putting x=-2 gives B=-7.

\displaystyle\int \dfrac{3x-1}{(x+2)^2}\,dx = 3\log|x+2| + \dfrac{7}{x+2} + C

Integrate: \dfrac{1}{x^4-1}

Hard +

Solution

Factor fully: x^4-1=(x-1)(x+1)(x^2+1).

Putting x=1 and x=-1 gives A=\dfrac14 and B=-\dfrac14; matching the remaining coefficients gives C=0, D=-\dfrac12.

\displaystyle\int \dfrac{1}{x^4-1}\,dx = \dfrac14\log\left|\dfrac{x-1}{x+1}\right| - \dfrac12\tan^{-1}x + C

Integrate: \dfrac{1}{x(x^n+1)} (Hint: multiply by x^{n-1}, put x^n=t)

Hard +

Solution

Multiplying by x^{n-1} top and bottom gives \dfrac{x^{n-1}}{x^n(x^n+1)}.

Put t=x^n, so x^{n-1}dx=\dfrac{dt}{n}, reducing the integral to \dfrac1n\displaystyle\int \dfrac{dt}{t(t+1)}.

Decomposing \dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1} and integrating gives \dfrac1n\log\left|\dfrac{t}{t+1}\right|.

\displaystyle\int \dfrac{1}{x(x^n+1)}\,dx = \dfrac1n\log\left|\dfrac{x^n}{x^n+1}\right| + C

Integrate: \dfrac{\cos x}{(1-\sin x)(2-\sin x)} (Hint: put \sin x=t)

Medium +

Solution

Put t=\sin x, so dt=\cos x\,dx, reducing the integral to \displaystyle\int \dfrac{dt}{(1-t)(2-t)}. Decomposing gives \dfrac{1}{1-t}-\dfrac{1}{2-t}.

\displaystyle\int \dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,dx = \log\left|\dfrac{2-\sin x}{1-\sin x}\right| + C

Integrate: \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}

Hard +

Solution

Put y=x^2; the fraction \dfrac{(y+1)(y+2)}{(y+3)(y+4)} is improper in y, and long division followed by partial fractions reduces it to 1+\dfrac{2}{y+3}-\dfrac{6}{y+4}.

Substituting back y=x^2 and integrating term by term.

\displaystyle\int \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx = x + \dfrac{2}{\sqrt3}\tan^{-1}\left(\dfrac{x}{\sqrt3}\right) - 3\tan^{-1}\left(\dfrac{x}{2}\right) + C

Integrate: \dfrac{2x}{(x^2+1)(x^2+3)}

Easy +

Solution

Put y=x^2, so dy=2x\,dx, reducing the integral to \displaystyle\int \dfrac{dy}{(y+1)(y+3)}, which decomposes as \dfrac{1/2}{y+1}-\dfrac{1/2}{y+3}.

\displaystyle\int \dfrac{2x}{(x^2+1)(x^2+3)}\,dx = \dfrac12\log\left|\dfrac{x^2+1}{x^2+3}\right| + C

Integrate: \dfrac{1}{x(x^4-1)}

Hard +

Solution

Factor fully: x(x^4-1)=x(x-1)(x+1)(x^2+1).

Putting x=0,1,-1 gives A=-1, B=\dfrac14, C=\dfrac14; matching the remaining coefficients gives D=\dfrac12, E=0.

\displaystyle\int \dfrac{1}{x(x^4-1)}\,dx = -\log|x| + \dfrac14\log|x-1| + \dfrac14\log|x+1| + \dfrac14\log(x^2+1) + C

Integrate: \dfrac{1}{e^x-1} (Hint: put e^x=t)

Medium +

Solution

Put t=e^x, so dx=\dfrac{dt}{t}, reducing the integral to \displaystyle\int \dfrac{dt}{t(t-1)}, which decomposes as \dfrac{1}{t-1}-\dfrac1t.

\displaystyle\int \dfrac{1}{e^x-1}\,dx = \log\left|\dfrac{e^x-1}{e^x}\right| + C

MCQ. \displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)} equals: (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

Easy +

Solution

Write \dfrac{x}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}: putting x=1 gives A=-1, and putting x=2 gives B=2.

So the integral is -\log|x-1|+2\log|x-2|.

Answer: (B) \log\left|\dfrac{(x-2)^2}{x-1}\right|+C

MCQ. \displaystyle\int \dfrac{dx}{x(x^2+1)} equals: (A) \log|x|-\dfrac12\log(x^2+1)+C (B) \log|x|+\dfrac12\log(x^2+1)+C (C) -\log|x|+\dfrac12\log(x^2+1)+C (D) \dfrac12\log|x|+\log(x^2+1)+C

Easy +

Solution

Write \dfrac{1}{x(x^2+1)} = \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}.

Putting x=0 gives A=1; comparing the x^2 coefficient gives B=-1; comparing the x coefficient gives C=0.

Answer: (A) \log|x|-\dfrac12\log(x^2+1)+C

Common Questions

FAQs: Chapter 7 Ex 7.5

How many questions are there in Class 12 Maths Chapter 7 Ex 7.5?

Exercise 7.5 has 23 questions in total — 21 rational functions to be integrated using partial fractions, followed by 2 MCQs.

What technique does Chapter 7 Ex 7.5 test?

It tests integration by partial fractions — decomposing a rational function into simpler fractions (distinct linear factors, repeated linear factors, or irreducible quadratic factors) that can each be integrated using standard formulae.

Where can I find the official NCERT textbook for this exercise?

The official NCERT Class 12 Maths textbook, including Chapter 7 (Integrals) and Exercise 7.5, is available for free at ncert.nic.in.

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.5 – Integration by Partial Fractions

Chapter 7 Ex 7.5 Solutions — All 23 Questions

Integrate: \dfrac{x}{(x+1)(x+2)}

Integrate: \dfrac{1}{x^2-9}

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

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

Integrate: \dfrac{2x}{x^2+3x+2}

Integrate: \dfrac{1-x^2}{x(1-2x)}

Integrate: \dfrac{x}{(x^2+1)(x-1)}

Integrate: \dfrac{x}{(x-1)^2(x+2)}

Integrate: \dfrac{3x+5}{x^3-x^2-x+1}

Integrate: \dfrac{2x-3}{(x^2-1)(2x+3)}

Integrate: \dfrac{5x}{(x+1)(x^2-4)}

Integrate: \dfrac{x^3+x+1}{x^2-1}

Integrate: \dfrac{2}{(1-x)(1+x^2)}

Integrate: \dfrac{3x-1}{(x+2)^2}

Integrate: \dfrac{1}{x^4-1}

Integrate: \dfrac{1}{x(x^n+1)} (Hint: multiply by x^{n-1}, put x^n=t)

Integrate: \dfrac{\cos x}{(1-\sin x)(2-\sin x)} (Hint: put \sin x=t)

Integrate: \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}

Integrate: \dfrac{2x}{(x^2+1)(x^2+3)}

Integrate: \dfrac{1}{x(x^4-1)}

Integrate: \dfrac{1}{e^x-1} (Hint: put e^x=t)

MCQ. \displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)} equals: (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

MCQ. \displaystyle\int \dfrac{dx}{x(x^2+1)} equals: (A) \log|x|-\dfrac12\log(x^2+1)+C (B) \log|x|+\dfrac12\log(x^2+1)+C (C) -\log|x|+\dfrac12\log(x^2+1)+C (D) \dfrac12\log|x|+\log(x^2+1)+C

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FAQs: Chapter 7 Ex 7.5

How many questions are there in Class 12 Maths Chapter 7 Ex 7.5?

What technique does Chapter 7 Ex 7.5 test?

Where can I find the official NCERT textbook for this exercise?

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