Question

$\underset{2}{\overset{4}{\int }}\frac{x}{x^2+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}{x}^2=t.\mathrm{Then},2xdx=dt \\ \mathrm{When}x=2,t=4\mathrm{and}x=4,t=16. \\ \therefore I={\int }_2^4\frac{x}{x^2+1}\mathit{}\mathit{d}x \\ \Rightarrow I={\int }_4^{16}\frac{1}{2}\frac{dt}{t+1} \\ \Rightarrow I=\frac{1}{2}{\left[\log \left(t+1\right)\right]}_4^{16} \\ \Rightarrow I=\frac{1}{2}\log 17-\frac{1}{2}\log 5 \\ \Rightarrow I=\frac{1}{2}\log \frac{17}{5} \end{gathered}$$