Evaluate: $\int \frac{2\log x+3}{x(3\log x+2)\left[(\log x)^2+1\right]}\,dx$

Evaluate: ∫ (2log x+3)/(x(3log x+2)[(log x)²+1])dx

3.4 · Q171

\int \frac{2\log x+3}{x(3\log x+2)\left[(\log x)^2+1\right]}\,dx

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Let $t=\log x$ , $dt=dx/x$ : the integral becomes $\int\dfrac{2t+3}{(3t+2)(t^2+1)}dt$ . Write $\dfrac{2t+3}{(3t+2)(t^2+1)}=\dfrac{A}{3t+2}+\dfrac{Bt+C}{t^2+1}$ , so $2t+3=A(t^2+1)+(Bt+C)(3t+2)$ . Put $t=-\frac23$ : $\frac53=\frac{13}{9}A \Rightarrow A=\frac{15}{13}$ . Comparing $t^2$ coefficients: $0=A+3B \Rightarrow B=-\frac{5}{13}$ . Comparing constants: $3=A+2C \Rightarrow C=\frac{12}{13}$ (check the t coefficient: $2=2B+3C=-\frac{10}{13}+\frac{36}{13}=2$ ). So $\int\frac{2t+3}{(3t+2)(t^2+1)}dt = \frac{A}{3}\log|3t+2| + \frac{B}{2}\log(t^2+1 …

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

…

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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 …

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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 …
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Q.Evaluate: $\int \frac{2\log x+3}{x(3\log x+2)\left[(\log x)^2+1\right]}\,dx$

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Q.Evaluate: $\int \frac{2\log x+3}{x(3\log x+2)\left[(\log x)^2+1\right]}\,dx$

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