Question

$\int \frac{x}{\sqrt{8+x-x^2}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{xdx}{\sqrt{8+x-x^2}} \\ \mathrm{Consider},x=A\frac{d}{dx}\left(8+x-x^2\right)+B \\ x=A\left(1-2x\right)+B \\ x=\left(-2A\right)x+A+B \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ -2A=1 \\ \Rightarrow A=-\frac{1}{2} \\ \mathrm{And} \\ A+B=0 \\ \Rightarrow -\frac{1}{2}+B=0 \\ \Rightarrow B=\frac{1}{2} \\ \therefore x=-\frac{1}{2}\left(1-2x\right)+\frac{1}{2} \\ \mathrm{Then}, \\ I=-\frac{1}{2}\int \frac{\left(1-2x\right)dx}{\sqrt{8+x-x^2}}+\frac{1}{2}\int \frac{dx}{\sqrt{8+x-x^2}} \\ =-\frac{1}{2}\int \frac{\left(1-2x\right)dx}{\sqrt{8+x-x^2}}+\frac{1}{2}\int \frac{dx}{\sqrt{8-\left(x^2-x\right)}} \\ =-\frac{1}{2}\int \frac{\left(1-2x\right)dx}{\sqrt{8+x-x^2}}+\frac{1}{2}\int \frac{dx}{\sqrt{8-\left(x^2-x+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}\right)}} \\ =-\frac{1}{2}\int \frac{\left(1-2x\right)dx}{\sqrt{8+x-x^2}}+\frac{1}{2}\int \frac{dx}{\sqrt{8+{\displaystyle \frac{1}{4}}-{\left(x-{\displaystyle \frac{1}{2}}\right)}^2}} \\ =-\frac{1}{2}\int \frac{\left(1-2x\right)dx}{\sqrt{8+x-x^2}}+\frac{1}{2}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\sqrt{33}}{2}}\right)}^2-{\left(x-{\displaystyle \frac{1}{2}}\right)}^2}} \\ \mathrm{let}8+x-x^2=t \\ \Rightarrow \left(1-2x\right)dx=dt \\ \therefore I=-\frac{1}{2}\int \frac{dt}{\sqrt{t}}+\frac{1}{2}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\sqrt{33}}{2}}\right)}^2-{\left(x-{\displaystyle \frac{1}{2}}\right)}^2}} \\ =-\frac{1}{2}\times 2\sqrt{t}+\frac{1}{2}{\sin }^{-1}\left(\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{33}}{2}}}\right)+C \\ =-\sqrt{t}+\frac{1}{2}{\sin }^{-1}\left(\frac{2x-1}{\sqrt{33}}\right)+C \\ =-\sqrt{8+x-x^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{2x-1}{\sqrt{33}}\right)+C \end{gathered}$$