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KCET · Maths · Indefinite Integration

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

  1. A \(2\left[x \tan { }^{1} x-\log \left(1+x^{2}\right)\right]+C\)
  2. B \(2\left[x \tan ^{1} x+\log \left(1+x^{2}\right)\right]+C\)
  3. C \(2 x \tan ^{-1} x+\log \left(1+x^{2}\right)+C\)
  4. D \(2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)

Step-by-step Solution

Detailed explanation

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