Question

$\int \frac{x}{\sqrt{9+8x-x^2}}dx$ is equal to

• A. $-\sqrt{9+8x-x^2}+4{\sin }^{-1}\left(\frac{x-4}{5}\right)+C$
• B. $-\sqrt{9+8x-x^2}+4{\cos }^{-1}\left(\frac{x-4}{5}\right)+C$
• C. $-\sqrt{9+8x-x^2}+4{\cos }^{-1}\left(\frac{x-3}{2}\right)+C$
• D. $\sqrt{9+8x-x^2}+4{\sin }^{-1}\left(\frac{x-4}{5}\right)+C$

Answer 1

The correct option is A $-\sqrt{9+8x-x^2}+4{\sin }^{-1}\left(\frac{x-4}{5}\right)+C$

$I=\int \frac{x}{\sqrt{9+8x-x^2}}dx$
We can write $x=A(8-2x)+B$
On comparing $A=-\frac{1}{2}$, $B=4$
$\therefore I=\int -\frac{1}{2}\frac{(8-2x)}{\sqrt{9+8x-x^2}}dx+4\int \frac{dx}{\sqrt{9+8x-x^2}}$
Put $9+8x-x^2=t$ in the first integral
$\Rightarrow (8-2x)dx=dt$
$\therefore I=-\frac{1}{2}\int \frac{1}{\sqrt{t}}dt+4\int \frac{dx}{\sqrt{9-(x^2-8x)}}$
$=-\frac{1}{2}\frac{\sqrt{t}}{\frac{1}{2}}+4\int \frac{dx}{\sqrt{25-(x^2-8x+16)}}$
$=-\sqrt{9+8x-x^2}+4\int \frac{dx}{\sqrt{5^2-{(x-4)}^2}}$
$=-\sqrt{9+8x-x^2}+4{\sin }^{-1}\left(\frac{x-4}{5}\right)+C$