Question

Integrate the following functions.
$\int \frac{x+2}{\sqrt{4x-x^2}}dx.$

Answer 1

Let $x+2=A\frac{d}{dx}(4x-x^2)+B\Rightarrow x+2=A(4-2x)+B\Rightarrow x+2=-2Ax+4A+B$
On equating the coefficients of x and constant term on both sides, we get
$-2A=1\Rightarrow A=-\frac{1}{2}and4A+B=2\Rightarrow B=4\Rightarrow (x+2)=-\frac{1}{2}(4-2x)+4\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx=\int \frac{-\frac{1}{2}(4-2x)+4}{\sqrt{4x-x^2}}dx=-\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}dx+4\int \frac{dx}{\sqrt{4x-x^2}}$
Let $I_1=\int \frac{4-2x}{\sqrt{4x-x^2}}dxand{I}_2=\int \frac{dx}{\sqrt{4x-x^2}}$
Then $\int \frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}{I}_1+4I_2...........\left(i\right)$
Now, $I_1=\int \frac{4-2x}{\sqrt{4x-x^2}}dx$
Let $4x-x^2=t\Rightarrow (4-2x)dx=dt\Rightarrow I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C_1$
$=2\sqrt{4x-x^2}+C_1.........\left(ii\right)$
And $I_2=\int \frac{dx}{\sqrt{4x-x^2}}$
$[\because 4x-x^2=-(x^2-4x)=-(x-2{)}^2-4=\left[\right(2{)}^2-(x-2{)}^2]\therefore I_2=\int \frac{1}{\sqrt{(2{)}^2-(x-2{)}^2}}dx=si{n}^{-1}\left(\frac{x-2}{2}\right)+C_2....\left(iii\right)[\because \int \frac{dx}{\sqrt{a^2-x^2}}=si{n}^{-1}(\frac{x}{a}\left)\right]$
On substituting the values of $I_1$ and $I_2$ from Eqs. (ii)and (iii)in Eq. (i), we get

$\int \frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}\left[2\sqrt{4x-x^2}\right]+4si{n}^{-1}\left(\frac{x-2}{2}\right)+C(\because -\frac{1}{2}{C}_1+4C_2=C)=-\sqrt{4x-x^2}+4si{n}^{-1}\left(\frac{x-2}{2}\right)+C$