Question

evaluate the following integral
integral of x^3/1+x^8 dx

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^3}{1+x^8}dx \\ \mathrm{Putting}{x}^4=t \\ \Rightarrow 4x^{3}dx=dt \\ \Rightarrow x^{3}dx=\frac{1}{4}dt \\ \therefore I=\frac{1}{4}\int \frac{1}{1+t^2}dt \\ =\frac{1}{4}{\tan }^{-1}t+C \\ =\frac{1}{4}{\tan }^{-1}\left(x^4\right)+C \end{gathered}$$