Question

$\int \frac{3x+1}{\sqrt{5-2x-x^2}}dx$

Answer 1

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