Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{\left(x+1\right)\left(x^2+2x+2\right)} \\ =\int \frac{dx}{\left(x+1\right)\left[x^2+2x+1+1\right]} \\ =\int \frac{dx}{\left(x+1\right)\left[{\left(x+1\right)}^2+1\right]} \\ \mathrm{Putting}x+1=t \\ \Rightarrow dx=dt \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes} \\ I=\int \frac{dt}{t\left[t^2+1\right]} \\ =\int \frac{t·dt}{t^2\left(t^2+1\right)} \\ \mathrm{Again}\mathrm{putting}{t}^2=p \\ \Rightarrow 2tdt=dp \\ \Rightarrow tdt=\frac{dp}{2} \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes} \\ I=\frac{1}{2}\int \frac{dp}{p\left(p+1\right)} \\ =\frac{1}{2}\int \frac{dp}{p^2+p} \\ =\frac{1}{2}\int \frac{dp}{p^2+p+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}} \\ =\frac{1}{2}\int \frac{dp}{{\left(p+{\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{1}{2}\left[\frac{1}{2\times {\displaystyle \frac{1}{2}}}\log \left|\frac{p+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}{p+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right|\right]+C \\ =\frac{1}{2}\log \left|\frac{p}{p+1}\right|+C \\ =\frac{1}{2}\log \left|\frac{t^2}{t^2+1}\right|+C \\ =\frac{1}{2}\log \left|\frac{{\left(x+1\right)}^2}{{\left(x+1\right)}^2+1}\right|+C \\ =\log \sqrt{\left|\frac{{\left(x+1\right)}^2}{{\left(x+1\right)}^2+1}\right|}+C \\ =\log \left|\frac{x+1}{\sqrt{x^2+2x+2}}\right|+C \end{gathered}$$