Integral Calculus
86471
When \(x>0\), then \(\int \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x\) is
1 \(2\left[x \tan ^{-1} \mathrm{x}-\log \left(1+\mathrm{x}^{2}\right)\right]+\mathrm{c}\)
2 \(2\left[x \tan ^{-1} \mathrm{x}+\log \left(1+\mathrm{x}^{2}\right)\right]+\mathrm{c}\)
3 \(2 x \tan ^{-1} \mathrm{x}+\log \left(1+\mathrm{x}^{2}\right)+\mathrm{c}\)
4 \(2 x \tan ^{-1} \mathrm{x}-\log \left(1+\mathrm{x}^{2}\right)+\mathrm{c}\)
Explanation:
(D) : When, \(\mathrm{x}>0\)
\(I=\int \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x\)
\(\because 2 \tan ^{-1} \mathrm{x}=\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)\)
\(\therefore \quad \mathrm{I}=\int 2 \tan ^{-1} \mathrm{x} d \mathrm{x}=2 \int 1 \cdot \tan ^{-1} \mathrm{x} d \mathrm{x}\)
\(=2\left[\mathrm{x} \cdot \tan ^{-1} \mathrm{x}-\int \frac{1}{1+\mathrm{x}^{2}} \cdot \mathrm{x} \mathrm{dx}\right]\)
\(=2 \mathrm{x} \cdot \tan ^{-1} \mathrm{x}-\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}\)
\(=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+c\)