Question

$\int \frac{x}{\left(x^2+4\right)\sqrt{x^2+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{xdx}{\left(x^2+4\right)\sqrt{x^2+1}} \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \frac{dt}{\left(t+4\right)\sqrt{t+1}} \\ \mathrm{Again}\mathrm{Putting}t+1=p^2 \\ \Rightarrow t=p^2-1 \\ \Rightarrow dt=2pdp \\ I=\frac{1}{2}\int \frac{2pdp}{\left(p^2-1+4\right)p} \\ =\int \frac{dp}{p^2+3} \\ =\int \frac{dp}{p^2+{\left(\sqrt{3}\right)}^2} \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{p}{\sqrt{3}}\right)+\mathrm{C} \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{\mathrm{t}+1}}{\sqrt{3}}\right)+\mathrm{C} \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\sqrt{\frac{x^2+1}{3}}\right)+\mathrm{C} \end{gathered}$$