Q112 / 180 · KCET 2011 · Mathematics

When $x>0$, then $\int \cos ^{1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x$ is

  • A
    $2\left[x \tan { }^{1} x-\log \left(1+x^{2}\right)\right]+C$
  • B
    $2\left[x \tan ^{1} x+\log \left(1+x^{2}\right)\right]+C$
  • C
    $2 x \tan ^{-1} x+\log \left(1+x^{2}\right)+C$
  • Correct answer
    D
    $2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C$

Solution

Given, when $x>0$ $$ =\int \cos ^{1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x $$ $\left[\begin{array}{l}\because x>0 \\ \therefore 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\end{array}\right]$ $=\int 2 \tan ^{-1} x d x$ $=2 \int \frac{1}{\text { II }} \tan ^{-1} x d x$ $=2\left\{x \cdot \tan ^{-1} x-\int \frac{1}{1+x^{2}} \cdot x d x\right\}$ $=2 x \tan ^{-1} x-\int \frac{2 x}{1+x^{2}} d x$ $=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C$

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

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