Question

$\int \frac{x}{\left(x^2+4\right)\sqrt{x^2+9}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{xdx}{\left(x^2+4\right)\sqrt{x^2+9}} \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \frac{dt}{\left(t+4\right)\sqrt{t+9}} \\ \mathrm{Again}\mathrm{Putting}t+9=u^2 \\ \Rightarrow dt=2udu \\ \therefore I=\frac{1}{2}\int \frac{2udu}{\left(u^2-9+4\right)u} \\ =\int \frac{du}{u^2-5} \\ =\int \frac{du}{{\displaystyle u^2-{\left(\sqrt{5}\right)}^2}} \\ =\frac{1}{2\sqrt{5}}\log \left|\frac{u-{\displaystyle \sqrt{5}}}{u+\sqrt{5}}\right|+C \\ =\frac{1}{2\sqrt{5}}\log \left|\frac{\sqrt{t+9}-\sqrt{5}}{\sqrt{t+9}+\sqrt{5}}\right|+C \\ =\frac{1}{2\sqrt{5}}\log \left|\frac{\sqrt{x^2+9}-\sqrt{5}}{\sqrt{x^2+9}+\sqrt{5}}\right|+C \end{gathered}$$