Question

Evaluate: $\int \frac{x^2}{1+x^3}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^{2}dx}{1+x^3} \\ \mathrm{Putting}1+x^3=t \\ \Rightarrow 3x^{2}dx=dt \\ \Rightarrow x^{2}dx=\frac{dt}{3} \\ \therefore I=\frac{1}{3}\int \frac{dt}{t} \\ =\frac{1}{3}\ln \left|t\right|+C \\ =\frac{1}{3}\ln \left|1+x^3\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \mathit{}t\mathit{=}1+x^{\mathit{3}}\right) \end{gathered}$$