Question

$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(5x+3\right)dx}{\sqrt{x^2+4x+10}} \\ \mathrm{Consider}, \\ 5x+3=A\frac{d}{dx}\left(x^2+4x+10\right)+B \\ \Rightarrow 5x+3=A\left(2x+4\right)+B \\ \Rightarrow 5x+3=\left(2A\right)x+4A+B \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2A=5 \\ \Rightarrow A=\frac{5}{2} \\ \mathrm{And} \\ 4A+B=3 \\ \Rightarrow 4\times \frac{5}{2}+B=3 \\ \Rightarrow B=-7 \\ \therefore I=\frac{5}{2}\int \frac{{\displaystyle \left(2x+4\right)dx}}{\sqrt{x^2+4x+10}}-7\int \frac{dx}{\sqrt{x^2+4x+10}} \\ =\frac{5}{2}\int \frac{\left(2x+4\right)dx}{\sqrt{x^2+4x+10}}-7\int \frac{dx}{\sqrt{x^2+4x+4-4+10}} \\ =\frac{5}{2}\int \frac{\left(2x+4\right)dx}{\sqrt{x^2+4x+10}}-7\int \frac{dx}{\sqrt{{\left(x+2\right)}^2+{\left(\sqrt{6}\right)}^2}} \\ \mathrm{Putting},x^2+4x+10=t \\ \Rightarrow \left(2x+4\right)dx=dt \\ \mathrm{Then}, \\ I=\frac{5}{2}\int \frac{dt}{\sqrt{t}}-7\log \left|x+2+\sqrt{{\left(x+2\right)}^2+6}\right|+C \\ =\frac{5}{2}\int t^{-\frac{1}{2}}dt-7\log \left|x+2+\sqrt{x^2+4x+10}\right|+C \\ =\frac{5}{2}\times 2\sqrt{t}-7\log \left|x+2+\sqrt{x^2+4x+10}\right|+C \\ =5\sqrt{x^2+4x+10}-7\log \left|x+2+\sqrt{x^2+4x+10}\right|+C \end{gathered}$$