Question

$\int \frac{x+\sqrt{x+1}}{x+2}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{x+\sqrt{x+1}}{x+2}dx \\ \mathrm{Let},x+1=t^2 \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get} \\ dx=2tdt \\ \mathrm{Now},\mathrm{integration}\mathrm{becomes} \\ I=\int \frac{\left(t^2-1+t\right)}{t^2+1}2tdt \\ =2\int \frac{t^3+t^2-t}{t^2+1}dt \\ =2\int \frac{t^3+t-t+t^2+1-1-t}{t^2+1}dt \\ =2\int \frac{t^3+t+t^2+1-t-t-1}{t^2+1}dt \\ =2\int \frac{t^3+t}{t^2+1}dt++2\int \frac{t^2+1}{t^2+1}dt+2\int \frac{-2t-1}{t^2+1}dt \\ =2\int tdt+2\int dt-2\int \frac{2t}{t^2+1}dt-2\int \frac{1}{t^2+1}dt \\ =t^2+2t-2\log \left|t^2+1\right|-2{\tan }^{-1}t+C \\ =\left(x+1\right)+2\sqrt{x+1}-2\log \left|x+2\right|-2{\tan }^{-1}\sqrt{x+1}+C \end{gathered}$$