Q110 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

$$ \int \frac{1}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} d x= $$

  • Correct answer
    A
    $\frac{1}{2} \sec ^{-1}\left(\mathrm{x}^2+2\right)+\mathrm{c}$
  • B
    $-\operatorname{cosec~} \mathrm{h}^{-1}\left(\mathrm{x}^2+2\right)+\mathrm{c}$
  • C
    $\frac{1}{2} \tan ^{-1}\left(x+\frac{2}{x}\right)+c$
  • D
    $-\frac{1}{2} \cot ^{-1}\left(x+\frac{2}{x}\right)+c$

Solution

$\begin{aligned} & \text {} \int \frac{d x}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} \\ & =\int \frac{x d x}{\left(x^2+2\right) \sqrt{\left(x^2+2\right)^2-1}}\end{aligned}$ $\begin{gathered}x^2+2=t \Rightarrow 2 x d x=d t \\ =\frac{1}{2} \int \frac{d t}{t \sqrt{t^2-1}}=\frac{1}{2} \sec ^{-1} t+C=\frac{1}{2} \sec ^{-1}\left(x^2+2\right)+C .\end{gathered}$

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

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