Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(2x+1\right)dx}{\sqrt{x^2+2x-1}} \\ =\int \frac{\left(2x+2-1\right)dx}{\sqrt{x^2+2x-1}} \\ =\int \frac{\left(2x+2\right)dx}{\sqrt{x^2+2x-1}}-\int \frac{dx}{\sqrt{x^2+2x-1}} \\ =\int \frac{\left(2x+2\right)dx}{\sqrt{x^2+2x-1}}-\int \frac{dx}{\sqrt{x^2+2x+1-1-1}} \\ =\int \frac{\left(2x+2\right)dx}{\sqrt{x^2+2x-1}}-\int \frac{dx}{\sqrt{{\left(x+1\right)}^2-{\left(\sqrt{2}\right)}^2}} \\ \mathrm{let}{x}^2+2x-1=t \\ \Rightarrow \left(2x+2\right)dx=dt \\ I=\int \frac{dt}{\sqrt{t}}-\int \frac{dx}{\sqrt{{\left(x+1\right)}^2-{\left(\sqrt{2}\right)}^2}} \\ =2\sqrt{t}-\log \left|x+1+\sqrt{{\left(x+1\right)}^2-{\left(\sqrt{2}\right)}^2}\right|+C \\ =2\sqrt{x^2+2x-1}-\log \left|x+1+\sqrt{x^2+2x-1}\right|+C \end{gathered}$$