Q111 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

$\int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x=$

  • A
    $\frac{-\sqrt{x^2-1}}{x}$
  • B
    $\frac{-\sqrt{x^2-1}}{x}$
  • C
    $\frac{-x}{\sqrt{x^2-1}}$
  • Correct answer
    D
    $\frac{-\sqrt{x^2-1}}{x^2}$

Solution

$u=\sqrt{x^2-1}$ $d u=\frac{1}{\not z}$ $\left(x^2-1\right)^{1 / 2-1} \not2 x$ $d_x$ $=\frac{x}{\sqrt{x^2-1}} d x$ $d x=\frac{\sqrt{x^2-1}}{x} d u$ $\begin{aligned} & u=\sqrt{x^2-1} \\ & x^2-1=u^2\end{aligned}$ $\begin{aligned} & x^2=u^2+1 \\ & x^4=\left(u^2+1\right)^2\end{aligned}$ $\int \frac{x^2-1-1}{x^3 \not y}$ $\frac{\sqrt{\not x-1}}{x} d u$ $\int \frac{x^2-1-1}{x^4} d u$ $\int \frac{u^2-1}{\left(u^2+1\right)^2} d u$ $\int \frac{u^2+1-2}{\left(u^2+1\right)^2} d u$ $\int \frac{u^2+1}{\left(u^2+1\right)^2} d u-\int \frac{2}{\left(u^2+1\right)^2} d u$ $\int \frac{d u}{u^2+1}-2 \int \frac{1}{\left(u^2+1\right)^2}$ $\int \frac{1}{\left(a x^2+b\right)^n} \Rightarrow \frac{2 n-3}{2 b(n-1)} \int \frac{1}{\left(a x^2+b\right)^{n-1}} d x+\frac{x}{2 b(n-1)\left(a x^2+b\right)^{n-1}}$ $\int \frac{1}{\left(u^2+1\right)^2}=\frac{1}{2(2-1)} \int \frac{1}{\left(u^2+1\right)} d u+\frac{u}{2(2-1)\left(u^2+1\right)}$ $=\frac{1}{2} \tan ^{-1}(u)+\frac{u}{2\left(v^2+1\right)}$ $\int \frac{d u}{u^2+1}-2 \int \frac{1}{\left(u^2+1\right)^2}$ $\tan ^{-1} u-2\left(\frac{1}{2} \tan ^{-1} u+\frac{v}{2\left(v^2+1\right)}\right)$ $=\tan ^{-1}(v)-\tan ^{-1}(u)-\frac{v}{\left(u^2+1\right)}$ $=\frac{-u}{u^2+1}$ $=-\frac{\sqrt{x^2-1}}{x^2}$

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

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