A) \[{{[{{\tan }^{-1}}{{x}^{2}}]}^{2}}+c\]
B) \[\frac{1}{2}{{[{{\tan }^{-1}}{{x}^{2}}]}^{2}}+c\]
C) \[2{{[{{\tan }^{-1}}{{x}^{2}}]}^{2}}+c\]
D) None of these
Correct Answer: B
Solution :
Put \[t={{\tan }^{-1}}{{x}^{2}}\Rightarrow dt=\frac{1}{1+{{x}^{4}}}\,2x\,dx,\] then \[\int_{{}}^{{}}{\frac{2x{{\tan }^{-1}}{{x}^{2}}}{1+{{x}^{4}}}}\,dx=\int_{{}}^{{}}{t\,dt}=\frac{{{t}^{2}}}{2}+c=\frac{1}{2}{{({{\tan }^{-1}}{{x}^{2}})}^{2}}+c.\]
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