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