Question

$\int \frac{x^3}{{\left(1+x^2\right)}^2}dx$

Answer 1

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