Q112 / 160 · AP EAMCET 2023 (18 May Shift 1) · Mathematics

$\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=$

  • A
    $\frac{1}{4} \log \left|\frac{x+1}{x-1}\right|-\frac{1}{2} \operatorname{Tan}^{-1} x+c$
  • Correct answer
    B
    $\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c$
  • C
    $\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \operatorname{Tan}^{-1} x+c$
  • D
    $\frac{1}{4} \log \left|\frac{x+1}{x-1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c$

Solution

Let $\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=I$ $\Rightarrow I=\frac{1}{2} \int\left(\frac{1}{x^2-1}+\frac{1}{x^2+1}\right) d x$ $\begin{aligned} & =\frac{1}{2} \cdot \frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c \\ & =\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c\end{aligned}$

Hence, the correct answer is option: 'B'.

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