A) \[-x+1(1+{{x}^{2}})ta{{n}^{-1}}x+c\]
B) \[x-(1+{{x}^{2}})co{{t}^{-1}}x+c\]
C) \[-x+(1+{{x}^{2}})co{{t}^{-1}}x+c\]
D) \[x-(1+{{x}^{2}})ta{{n}^{-1}}x+c\]
Correct Answer: A
Solution :
Let\[I=\int_{{}}^{{}}{xco{{s}^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)dx}\] \[\therefore \]\[I=2\int_{{}}^{{}}{\underset{II}{\mathop{x}}\,\underset{I}{\mathop{{{\tan }^{-1}}}}\,}xdx\] Applying Integration by parts \[I=2\left[ {{\tan }^{-1}}x\int_{{}}^{{}}{xdx-}\int_{{}}^{{}}{\left( \frac{d}{dx}(ta{{n}^{-1}}x \right)\int_{{}}^{{}}{xdx)dx}} \right]\] \[I=2\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\int_{{}}^{{}}{\frac{1}{1+{{x}^{2}}}\times \frac{{{x}^{2}}}{2}dx} \right]+c\] \[I=2\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\int_{{}}^{{}}{\frac{{{x}^{2}}+1-1}{1+{{x}^{2}}}dx} \right]+c\] \[I=2\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{\frac{\cancel{{{x}^{2}}+1}}{\cancel{{{x}^{2}}+1}}dx+\frac{1}{2}}\int_{{}}^{{}}{\frac{1}{1+{{x}^{2}}}dx} \right]+c\]\[I=2\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{1.dx+\frac{1}{2}{{\tan }^{-1}}x} \right]+c\] \[I=2\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}+\frac{1}{2}{{\tan }^{-1}}x \right]+c\] \[I={{x}^{2}}{{\tan }^{-1}}x+{{\tan }^{-1}}x-x+c\] or\[\]
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