A) \[{{({{\tan }^{-1}}x)}^{4}}+c\]
B) \[\frac{{{({{\tan }^{-1}}x)}^{4}}}{4}+c\]
C) \[2{{\tan }^{-1}}x+c\]
D) \[2{{({{\tan }^{-1}}x)}^{2}}+c\]
Correct Answer: B
Solution :
Put \[{{\tan }^{-1}}x=t\Rightarrow \frac{1}{1+{{x}^{2}}}dx=dt\] \ \[\int_{{}}^{{}}{\frac{{{({{\tan }^{-1}}x)}^{3}}}{1+{{x}^{2}}}}d\alpha =\int_{{}}^{{}}{{{t}^{3}}}dt=\frac{{{t}^{4}}}{4}+c\] = \[\frac{{{({{\tan }^{-1}}x)}^{4}}}{4}+c\].
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