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