Question

$\int \frac{x}{\left(x-3\right)\sqrt{x+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{xdx}{\left(x-3\right)\sqrt{x+1}} \\ \mathrm{Putting}x+1=t^2 \\ \Rightarrow x=t^2-1 \\ \mathrm{Diff}\mathrm{both}\mathrm{sides} \\ dx=2tdt \\ \therefore I=\int \frac{\left(t^2-1\right)2tdt}{\left(t^2-1-3\right)t} \\ =2\int \left(\frac{t^2-1}{t^2-4}\right)dt \\ =2\int \left(\frac{t^2-4+3}{t^2-4}\right)dt \\ =2\int \left(\frac{t^2-4}{t^2-4}\right)dt+6\int \frac{dt}{t^2-2^2} \\ =2\int dt+6\int \frac{dt}{t^2-2^2} \\ =2t+6\times \frac{1}{2\times 2}\log \left|\frac{t-2}{t+2}\right|+\mathrm{C} \\ =2\sqrt{x+1}+\frac{3}{2}\log \left|\frac{t-2}{t+2}\right|+\mathrm{C} \\ =2\sqrt{x+1}+\frac{3}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+\mathrm{C} \end{gathered}$$