CUET Preparation Today
| The value of \(\int \frac{x^{2}\tan^{-1}(x^{3})}{1+x^{6}}dx\) is |
\(\tan^{-1}(x^{3})+c\) \(\frac{1}{6}(\tan^{-1}x^{3})^{2}+c\) \(\frac{1}{2}\tan^{-1}x^{3}\) \(\frac{1}{2}(\tan^{-1}x^{3})^{3})+c\) |
| \(\frac{1}{6}(\tan^{-1}x^{3})^{2}+c\) |
| Let \(t=\tan^{-1}x^{3}\) then \(\begin{aligned}\frac{x^{2}\tan^{-1}(x^{3})}{1+x^{6}}dx&=\frac{1}{3}\int tdt\\ &=\frac{1}{6}(\tan^{-1}x^{3})^{2}+c\end{aligned}\) |
