Q73 / 125 · BITSAT 2013 · Mathematics

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

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

Solution

$\mathrm{I}=\int \frac{\mathrm{x}^{2}\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}}{\mathrm{x}^{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)^{1 / 2}}$ Let $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{p} \Rightarrow\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}=\mathrm{dp}$ $\mathrm{I}=\int \frac{\mathrm{dp}}{\mathrm{p} \sqrt{\mathrm{p}^{2}-2}}=\frac{1}{\sqrt{2}} \mathrm{sec}^{-1} \frac{\mathrm{p}}{\sqrt{2}}$ $=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{\mathrm{x}^{2}+1}{\sqrt{2} \mathrm{x}}\right)+\mathrm{c}$

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

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