Question

Integrate the function $\frac{x+2}{\sqrt{4x-x^2}}$

Answer 1

Let $x+2=A\frac{d}{dx}(4x-x^2)+B$
$\Rightarrow x+2=A(4-2x)+B$
Equating the coefficients of $x$ and constant term on both sides, we obtain
$-2A=1\Rightarrow A=\frac{1}{2}$
& $4A+B=2\Rightarrow B=4$
$\Rightarrow (x+2)=-\frac{1}{2}(4-2x)+4$
$\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx=\int \frac{-\frac{1}{2}(4-2x)+4}{\sqrt{4x-x^2}}dx$
$=-\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}dx+4\int \frac{1}{\sqrt{4x-x^2}}dx$
Let $I_1=\int \frac{4-2x}{\sqrt{4x-x^2}}dx$ and $I_2=\int \frac{1}{\sqrt{4x-x^2}}dx$
$\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}{I}_1+4I_2$ ........ (1)
Then, $I_1=\int \frac{4-2x}{\sqrt{4x-x^2}}dx$
Let $4x-x^2=t$
$\Rightarrow (4-2x)dx=dt$
$\Rightarrow I_1=\int \frac{1}{\sqrt{t}}dt=2\sqrt{t}=2\sqrt{4x-x^2}$ .......... (2)
$I_2=\int \frac{1}{\sqrt{4x-x^2}}dx$
$\Rightarrow 4x-x^2=-(-4x+x^2)$
$=(-4x+x^2+4-4)$
$=4-(x-2{)}^2$
$=(2{)}^2-(x-2{)}^2$
$\therefore I_2=\int \frac{1}{\sqrt{(2{)}^2-(x-2{)}^2}}dx={\sin }^{-1}\left(\frac{x-2}{2}\right)$ .......... (3)
Using equations (2) and (3) in (1), we obtain
$\int \frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}\left(2\sqrt{4x-x^2}\right)+4{\sin }^{-1}\left(\frac{x-2}{2}\right)+C$
$=-\sqrt{4x-x^2}+4{\sin }^{-1}\left(\frac{x-2}{2}\right)+C$