Question

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

Answer 1

​$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(x+2\right)dx}{\sqrt{x^2+2x-1}} \\ \mathrm{Consider}, \\ x+2=A\frac{d}{dx}\left(x^2+2x-1\right)+B \\ \Rightarrow x+2=A\left(2x+2\right)+B \\ \Rightarrow x+2=\left(2A\right)x+2A+B \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2A=1 \\ \Rightarrow A=\frac{1}{2} \\ \mathrm{And} \\ 2A+B=2 \\ \Rightarrow 2\times \frac{1}{2}+B=2 \\ \Rightarrow B=1 \\ Then, \\ I=\int \frac{\left[{\displaystyle \frac{1}{2}\left(2x+2\right)+1}\right]}{\sqrt{x^2+2x-1}}dx \\ =\frac{1}{2}\int \frac{\left(2x+2\right)dx}{\sqrt{x^2+2x-1}}+\int \frac{dx}{\sqrt{x^2+2x-1}} \\ \mathrm{let}{x}^2+2x-1=t \\ \Rightarrow \left(2x+2\right)dx=dt \\ \therefore I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}+\int \frac{dx}{\sqrt{x^2+2x-1}} \\ =\frac{1}{2}\int t^{-\frac{1}{2}}dt+\int \frac{dx}{\sqrt{x^2+2x+1-2}} \\ =\frac{1}{2}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+\int \frac{dx}{\sqrt{{\left(x+1\right)}^2-{\left(\sqrt{2}\right)}^2}} \\ =\sqrt{t}+\log \left|x+1+\sqrt{{\left(x+1\right)}^2-{\left(\sqrt{2}\right)}^2}\right|+C \\ =\sqrt{x^2+2x-1}+\log \left|x+1+\sqrt{x^2+2x-1}\right|+C \end{gathered}$$