Question

$\int \frac{1}{(x^2-6x+9)\sqrt{x^2-6x+4}}dx$ is equal to

• A. $\frac{1}{5}.\frac{x^2-6x+4}{(x-3)}+C$
• B. $\frac{2}{3}.\frac{\sqrt{x^2-6x+9}}{(x-5)}+C$
• C. $\frac{1}{5}.\frac{\sqrt{x^2-6x+4}}{(x-3)}+C$
• D. $\frac{1}{5}.\frac{\sqrt{x^2-6x+9}}{(x-2)}+C$

Answer 1

The correct option is C $\frac{1}{5}.\frac{\sqrt{x^2-6x+4}}{(x-3)}+C$

$I=\int \frac{1}{(x^2-6x+9)\sqrt{x^2-6x+4}}dx$
$=\int \frac{dx}{{(x-3)}^2\sqrt{{(x-3)}^2-5}}$
Put $x-3=\frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^2}dt$
$=\int \frac{-\frac{1}{t^2}dt}{\frac{1}{t^2}\sqrt{\frac{1}{t^2}-5}}$
$=-\int \frac{tdt}{\sqrt{1-5t^2}}$
Put $1-5t^2=u$
$\Rightarrow 10tdt=du$
$I=\frac{2}{10}\sqrt{1-5t^2}+C$
$=\frac{1}{5}.\frac{\sqrt{x^2-6x+4}}{(x-3)}+C$