Evaluate: $\int \frac{x^2}{(x^2+1)(x^2-2)(x^2+3)}\,dx$

Evaluate: ∫ (x²)/((x²+1)(x²-2)(x²+3))dx

3.4 · Q150

\int \frac{x^2}{(x^2+1)(x^2-2)(x^2+3)}\,dx

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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✓ Free question

Treat the denominator as a product of three factors linear in $m=x^2$ : $\dfrac{m}{(m+1)(m-2)(m+3)} = \dfrac{A}{m+1}+\dfrac{B}{m-2}+\dfrac{C}{m+3}$ , so $m = A(m-2)(m+3)+B(m+1)(m+3)+C(m+1)(m-2)$ . Put $m=-1$ : $-1=-6A \Rightarrow A=\frac16$ . Put $m=2$ : $2=15B \Rightarrow B=\frac{2}{15}$ . Put $m=-3$ : $-3=10C \Rightarrow C=-\frac{3}{10}$ . This is an algebraic device only (not a calculus substitution), so replace $m$ by $x^2$ and integrate directly in $x$ : $\int\frac{x^2\,dx}{(x^2+1)(x^2-2)(x^2+3)} = \frac16\int\frac{dx}{x^2+1} + \frac{2}{15}\int\frac{dx}{x^2-2} - \frac{3}{10}\int\frac{dx}{x^2+3}$ . Using $\int\frac{dx}{x^2+1}=\tan^{-1}x$ , $\int\frac{dx}{x^2-2}=\frac{1}{2\sqrt2}\log\left|\frac{x-\sqrt2}{x+\sqrt2}\right|$ , $\int\frac{dx}{x^2+3}=\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}$ , and simplifying the constants $\frac{2}{15}\cdot\frac{1}{2\sqrt2}=\frac{\sqrt2}{30}$ and $\frac{3}{10}\cdot\frac{1}{\sqrt3}=\frac{\sqrt3}{10}$ gives the result.

✓Final answer

\frac{1}{6}\tan^{-1}x + \frac{\sqrt2}{30}\log\left|\frac{x-\sqrt2}{x+\sqrt2}\right| - \frac{\sqrt3}{10}\tan^{-1}\frac{x}{\sqrt3} + c

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$\int \frac{x}{(1 + x)(2 + x)}\,dx = \int \left(\frac{A}{1 + x} + \frac{B}{2 + x}\right)dx$
›Reveal solutionSolution

Partial fractions give $A = -1,\; B = 2$ , so the claim $A = 1,\, B = 2$ is False.

Decompose the integrand:

$\frac{x}{(1 + x)(2 + x)} = \frac{A}{1 + x} + \frac{B}{2 + x}.$

Multiply through by $(1+x)(2+x)$ :

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

Put $x = -1$ : $\;-1 = A(2 - 1) + B(0) = A \Rightarrow A = -1.$

Put $x = -2$ : $\;-2 = A(0) + B(1 - 2) = -B \Rightarrow B = 2.$

So the true values are $A = -1$ and $B = 2$ . The statement claims $A = 1$ , which is wrong.

✓Final answer
$A = -1$
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{(x - 8)(x + 7)} =$
(a) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x - 1}\right| + c$
(b) $\dfrac{1}{15} \log \left|\dfrac{x + 8}{x + 7}\right| + c$
(c) $\dfrac{1}{15} \log \left|\dfrac{x - 8}{x + 7}\right| + c$
(d) $(x - 8)(x - 7) + c$
(e) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x + 1}\right| + c$
(f) $(x - 8)(x + 7) + c$
›Reveal solutionSolution

Partial fractions give $\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\!\left(\dfrac{1}{x-8} - \dfrac{1}{x+7}\right)$ , which integrates to $\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c$ .

Write

$\frac{1}{(x-8)(x+7)} = \frac{A}{x-8} + \frac{B}{x+7} \;\Rightarrow\; A(x+7) + B(x-8) = 1.$

Put $x = 8$ : $\;A(15) = 1 \Rightarrow A = \dfrac{1}{15}$ .

Put $x = -7$ : $\;B(-15) = 1 \Rightarrow B = -\dfrac{1}{15}$ .

Hence

$\int \frac{dx}{(x-8)(x+7)} = \frac{1}{15}\int\frac{dx}{x-8} - \frac{1}{15}\int\frac{dx}{x+7} = \frac{1}{15}\log|x-8| - \frac{1}{15}\log|x+7| + c.$

Combining the logs:

$= \frac{1}{15}\log\left|\frac{x-8}{x+7}\right| + c.$

✓Final answer
$\dfrac{1}{15}\log\left|\dfrac{x - 8}{x + 7}\right| + c$
CBSE 2022Set ANNUAL1 mark
$\int \dfrac{x}{(x + 2)(x + 3)}\, dx =$
›Reveal solutionSolution
Partial fractions give $\dfrac{x}{(x+2)(x+3)} = \dfrac{-2}{x+2} + \dfrac{3}{x+3}$ , so the blank preceding $\displaystyle\int \dfrac{3}{x+3}\,dx$ is $\displaystyle\int \dfrac{-2}{x+2}\,dx$ .

Resolve the integrand into partial fractions:

$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}.$

Clearing denominators: $x = A(x+3) + B(x+2)$ .

Put $x = -2$ : $-2 = A(1) \Rightarrow A = -2$ .

Put $x = -3$ : $-3 = B(-1) \Rightarrow B = 3$ .

Hence

$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3},$

so

$\int \frac{x}{(x+2)(x+3)}\,dx = \int \frac{-2}{x+2}\,dx + \int \frac{3}{x+3}\,dx.$

Comparing with the given form, the blank term is $\displaystyle\int \dfrac{-2}{x+2}\,dx = -2\log|x+2|$ .

✓Final answer
$\displaystyle\int \dfrac{-2}{x+2}\,dx = -2\log|x+2| \, (+c)$

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Concept understanding — Integration by Partial Fractions

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Q.Evaluate: $\int \frac{x^2}{(x^2+1)(x^2-2)(x^2+3)}\,dx$

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