Q72 / 170 · Manipal (MET) 2015 · Mathematics

$\int \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x$ is equal to

  • A
    $\sqrt{2} \sin ^{-1}\left\{\frac{\sqrt{2} x}{x^2+1}\right\}+C$
  • Correct answer
    B
    $\frac{1}{\sqrt{2}} \sin ^{-1}\left\{\frac{\sqrt{2} x}{x^2+1}\right\}+C$
  • C
    $\frac{1}{2} \sin ^{-1}\left\{\frac{\sqrt{2} x}{x^2+1}\right\}+C$
  • D
    None of these

Solution

Let $\begin{aligned} I & =\int \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\ & =\int \frac{\frac{1}{x^2}-1}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} d x\end{aligned}$ $\begin{aligned} & =-\int \frac{1}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-(\sqrt{2})^2}} d\left(x+\frac{1}{x}\right) \\ & =\frac{1}{\sqrt{2}} \operatorname{cosec}^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C \\ & =\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{\sqrt{2} x}{x^2+1}\right)+C\end{aligned}$

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

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