Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{xdx}{\left(x^2+2x+2\right)\sqrt{x+1}} \\ =\int \frac{xdx}{\left[{\left(x+1\right)}^2+1\right]\sqrt{x+1}} \\ \mathrm{Putting}x+1=t^2 \\ \Rightarrow x=t^2-1 \\ \mathrm{Diff}\mathrm{both}\mathrm{sides} \\ dx=2tdt \\ \therefore I=\int \frac{\left(t^2-1\right)2tdt}{\left[{\left(t^2\right)}^2+1\right]t} \\ =2\int \frac{\left(t^2-1\right)dt}{t^4+1} \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{t}^2 \\ I=2\left(\frac{1-{\displaystyle \frac{1}{t^2}}}{t^2+{\displaystyle \frac{1}{t^2}}}\right)dt \end{gathered}$$

$$\begin{gathered} =2\int \frac{\left(1-{\displaystyle \frac{1}{t^2}}\right)dt}{t^2+{\displaystyle \frac{1}{t^2}}+2-2} \\ =2\int \frac{\left(1-{\displaystyle \frac{1}{t^2}}\right)dt}{{\left(t+{\displaystyle \frac{1}{t}}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ \mathrm{Putting}t+\frac{1}{t}=p \\ \Rightarrow \left(1-\frac{1}{t^2}\right)dt=dp \\ I=2\int \frac{dp}{p^2-{\left(\sqrt{2}\right)}^2} \\ =2\times \frac{1}{2\sqrt{2}}\log \left|\frac{p-\sqrt{2}}{p+\sqrt{2}}\right|+C \\ =\frac{1}{\sqrt{2}}\log \left|\frac{p-\sqrt{2}}{P+\sqrt{2}}\right|+C \\ =\frac{1}{\sqrt{2}}\log \left|\frac{t+{\displaystyle \frac{1}{t}}-\sqrt{2}}{t+{\displaystyle \frac{1}{t}}+\sqrt{2}}\right|+C \\ =\frac{1}{\sqrt{2}}\log \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right|+C \\ =\frac{1}{\sqrt{2}}\log \left|\frac{x+1-\sqrt{2\left(x+1\right)}+1}{x+1+\sqrt{2\left(x+1\right)}+1}\right|+C \\ =\frac{1}{\sqrt{2}}\log \left|\frac{\left(x+2\right)-\sqrt{2\left(x+1\right)}}{\left(x+2\right)+\sqrt{2\left(x+1\right)}}\right|+C \end{gathered}$$