Question

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

• 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. $\frac{1}{2}(\tan^{- 1}x^{3})^{2} + C$

Answer 1

Answer: (B)

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

Answer 2

Put $\tan^{- 1}x^{3} = t$

$\frac{1}{1 + x^{6}}.3x^{2}dx = dt$ $\Rightarrow \frac{x^{2}}{1 + x^{6}}dx = \frac{dt}{3}$ $\mathbf{\Rightarrow}$ $\frac{1}{3}\int_{}^{}{tdt}$

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