Question

$\int \left(2x+3\right)\sqrt{x^2+4x+3}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(2x+3\right)\sqrt{x^2+4x+3}dx \\ \mathrm{Also},2x+3=\lambda \frac{d}{dx}\left(x^2+4x+3\right)+\mu \\ \Rightarrow 2x+3=\lambda \left(2x+4\right)+\mu \\ \Rightarrow 2x+3=\left(2\lambda \right)x+4\lambda +\mu \\ \mathrm{Equating}\mathrm{coefficient}\mathrm{of}\mathrm{like}\mathrm{terms}. \\ 2\lambda =2 \\ \Rightarrow \lambda =1 \\ \mathrm{And} \\ 4\lambda +\mu =3 \\ \Rightarrow 4+\mu =3 \\ \Rightarrow \mu =-1 \\ \therefore I=\int \left(2x+4-1\right)\sqrt{x^2+4x+3}dx \\ =\int \left(2x+4\right)\sqrt{x^2+4x+3}dx-\int \sqrt{x^2+4x+3}dx \\ =\int \left(2x+4\right)\sqrt{x^2+4x+3}dx-\int \sqrt{x^2+4x+4-1}dx \\ =\int \left(2x+4\right)\sqrt{x^2+4x+3dx}-\int \sqrt{{\left(x+2\right)}^2-1^2}dx \\ \mathrm{Let}{x}^2+4x+3=t \\ \Rightarrow \left(2x+4\right)dx=dt \\ \mathrm{Then}, \\ I=\int \sqrt{t}dt-\int \sqrt{{\left(x+2\right)}^2-1^2}dx \\ =\frac{2}{3}{t}^{\frac{3}{2}}-\left[\frac{x+2}{2}\sqrt{{\left(x+2\right)}^2-1}-\frac{1^2}{2}\log \left|\left(x+2\right)+\sqrt{{\left(x+2\right)}^2-1}\right|\right]+\mathrm{C} \\ =\frac{2}{3}{\left(x^2+4x+3\right)}^{\frac{3}{2}}-\frac{1}{2}\left[\left(x+2\right)\sqrt{x^2+4x+3}-\log \left|\left(x+2\right)+\sqrt{x^2+4x+3}\right|\right]+\mathrm{C} \end{gathered}$$