Question

$\int \frac{x}{\sqrt{(9+8x-x^2)}}dx.$

• A. $\sqrt{9+8x-x^2}-{\sin }^{-1}\frac{x-4}{5}$
• B. $-\sqrt{9+8x+x^2}+{\sin }^{-1}\frac{x-4}{5}$
• C. $-\sqrt{9+8x-x^2}+{\sin }^{-1}\frac{x-4}{5}$
• D. $-\sqrt{9-8x+x^2}+{\sin }^{-1}\frac{x-4}{5}$

Answer 1

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

Let $I=\int \frac{x}{\sqrt{(9+8x-x^2)}}dx$

Comparing $x=l(8-2x)+m\Rightarrow -2l=1,8l+m=0\Rightarrow I=-\frac{1}{2},m=4$

Therefore

$I=l\int \frac{8-2x}{\sqrt{(9+8x-x^2)}}dx+m\int \frac{dx}{\sqrt{[25-(x^2-8x+16)]}}$

Put $9+8x-x^2=t\Rightarrow (8-2x)dx=dt$

Therefore

$I=l\int \frac{1}{\sqrt{\left(t\right)}}dt+m\int \frac{dx}{\sqrt{[5^2-{(x-4)}^2]}}$
$=l\times 2\sqrt{t}+m{\sin }^{-1}\frac{x-4}{5}=-\sqrt{9+8x-x^2}+{\sin }^{-1}\frac{x-4}{5}$

Hence, option 'C' is correct.