\[\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}}{1+{{x}^{6}}}\ dx}\] is equal to [MP PET 1999; UPSEAT 1999]

\[-\frac{1}{2}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\]

\[{{\tan }^{-1}}({{x}^{3}})+c\]

\[\frac{1}{6}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\]

\[\frac{1}{2}{{({{\tan }^{-1}}{{x}^{2}})}^{3}}+c\]

\[\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}}{1+{{x}^{6}}}\

1.	\[\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}}{1+{{x}^{6}}}\ dx}\] is equal to [MP PET 1999; UPSEAT 1999]
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.	\[\frac{1}{2}{{({{\tan }^{-1}}{{x}^{2}})}^{3}}+c\]
Answer» C. \[-\frac{1}{2}{{({{\tan }^{-1}}{{x}^{3}})}^{2}}+c\]

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\[\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}}{1+{{x}^{6}}}\ dx}\] is equal to [MP PET 1999; UPSEAT 1999]

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