Question

$\underset{0}{\overset{1}{\int }}\frac{24x^3}{{\left(1+x^2\right)}^4}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{24x^3}{{\left(1+x^2\right)}^4}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}\mathit{,} \\ \mathrm{Let}{x}^2=t.\mathrm{Then},2xdx=dt \\ \mathrm{When}x=,t=0\mathrm{and}x=1,t=1 \\ \therefore I={\int }_0^1\frac{12t}{{\left(1+t\right)}^4}dt \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ I=12{\left[\frac{t}{-3{\left(1+t\right)}^3}\right]}_0^1+12{\int }_0^1\frac{1}{3{\left(1+t\right)}^3}dt \\ \Rightarrow I=12\left\{{\left[\frac{t}{-3{\left(1+t\right)}^3}\right]}_0^1-{\left[\frac{1}{6{\left(1+t\right)}^2}\right]}_0^1\right\} \\ \Rightarrow I=12\left\{-\frac{1}{24}-0-\frac{1}{24}+\frac{1}{6}\right\} \\ \Rightarrow I=12\times \frac{1}{12} \\ \Rightarrow I=1 \end{gathered}$$