Question

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

Answer 1

Let ${\displaystyle I=\int \frac{x+2}{\sqrt{4x-x^2}} dx}$

${\displaystyle =\int \frac{x+2}{\sqrt{-(x^2-4x+4)+4}} dx}$

${\displaystyle =\int \frac{x+2}{\sqrt{4-{(x-2)}^2}} dx}$

${\displaystyle =\frac{x-2+4}{\sqrt{4-{(x-2)}^2}} dx}$

${\displaystyle =\int \frac{x-2}{\sqrt{4-{(x-2)}^2}} dx+4 \int \frac{1}{\sqrt{4-{(x-2)}^2}} dx}$

Let ${\displaystyle I=I_1+4{\sin }^{-1}\left(\frac{x-2}{2}\right)+C_1}$          .....(i)

Where ${\displaystyle I_1=\int \frac{x-2}{\sqrt{4-{(x-2)}^2}}dx}$

Put 4 - (x - 2)² = t 

-2 (x - 2) dx = dt

${\displaystyle I_1=\frac{1}{2}\int \frac{dt}{\sqrt{{\left(2\right)}^2-t}}}$

${\displaystyle \frac{1}{2}\int \frac{dt}{\sqrt{4-t}}}$

${\displaystyle \frac{1}{2}\left[\frac{{(4-t)}^{-\frac{1}{2+1}}}{-(-\frac{1}{2}+1)}\right]+C_2}$

${\displaystyle =-\sqrt{(4-t)}+C_2}$

${\displaystyle =-\sqrt{4-{(x-2)}^2}+C_2}$

${\displaystyle =-\sqrt{4-x^2-4+4x}+C_2}$

${\displaystyle =-\sqrt{4x-x^2}+C_2}$                      ....(ii)

${\displaystyle I=-\sqrt{4x-x^2}+4{\sin }^{-1}\left(\frac{x-2}{2}\right)+C}$        ... [from (i) and (ii)]