Question

$\underset{1}{\overset{2}{\int }}\frac{3x}{9x^2-1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}{x}^2=t.\mathrm{Then},2xdx=dt \\ \mathrm{When}x=1,t=1\mathrm{and}x=2,t=4 \\ \therefore I={\int }_1^2\frac{3x}{9x^2-1}\mathit{}\mathit{d}x \\ \Rightarrow I=\frac{3}{2}{\int }_1^4\frac{dt}{9t-1} \\ \Rightarrow I=\frac{3}{18}{\left[\log \left(9t-1\right)\right]}_1^4 \\ \Rightarrow I=\frac{3}{18}\left(\log 35-\log 8\right) \\ \Rightarrow I=\frac{\left(\log 35-\log 8\right)}{6} \end{gathered}$$