indefinite-integration

Question: $\int_{{}}^{{}}{\frac{{{x}^{2}}{{\tan }^{-1}}{{x}^{3}}}{1+{{x}^{6}}}\ dx}$ is equal to

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

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

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

4) $\frac{1}{2}{{({{\tan }^{-1}}{{x}^{2}})}^{3}}+c$

Solution:

Explanation: No Explanation

Integration by substitution

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