Question

$\int \frac{2x+3}{\sqrt{x^2+4x+5}}dx$

Answer 1

​$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(2x+3\right)dx}{\sqrt{x^2+4x+5}} \\ =\int \frac{\left(2x+4-1\right)}{\sqrt{x^2+4x+5}}dx \\ =\int \frac{\left(2x+4\right)dx}{\sqrt{x^2+4x+5}}-\int \frac{dx}{\sqrt{x^2+4x+5}} \\ =\int \frac{\left(2x+4\right)dx}{\sqrt{x^2+4x+5}}-\int \frac{dx}{\sqrt{{\left(x+2\right)}^2+1}} \\ \mathrm{Consider}, \\ {x}^2+4x+5=t \\ \Rightarrow \left(2x+4\right)dx=dt \\ \therefore I=\int \frac{dt}{\sqrt{t}}-\int \frac{dx}{\sqrt{{\left(x+2\right)}^2+1^2}} \\ =\int t^{-\frac{1}{2}}dt-\int \frac{dx}{\sqrt{{\left(x+2\right)}^2+1^2}} \\ =\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}-\log \left|x+2+\sqrt{{\left(x+2\right)}^2+1}\right|+C \\ =2\sqrt{x^2+4x+5}-\log \left|x+2+\sqrt{x^2+4x+5}\right|+C \end{gathered}$$