A) \[{{\tan }^{-1}}({{x}^{3}})+c\]
B) \[\frac{1}{6}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\]
C) \[-\frac{1}{2}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\]
D) \[\frac{1}{2}{{({{\tan }^{-1}}{{x}^{2}})}^{3}}+c\]
Correct Answer: B
Solution :
Put \[{{x}^{3}}=t\Rightarrow dt=3{{x}^{2}}\,dx\] \[\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}\,dx}{1+{{x}^{6}}}}=\frac{1}{3}\int_{{}}^{{}}{\frac{{{\tan }^{-1}}t}{1+{{t}^{2}}}}\,dt\] Put \[z={{\tan }^{-1}}t\Rightarrow dz=\frac{dt}{1+{{t}^{2}}}\] \[=\frac{1}{3}\int_{{}}^{{}}{z\,dz}=\frac{1}{3}\frac{{{z}^{2}}}{2}=\frac{{{z}^{2}}}{6}=\frac{1}{6}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\].
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