Question

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

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

Answer 1

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

Let $I=\int \frac{dx}{(x^2-6x+9)\sqrt{x^2-6x+4}}$
$=\int \frac{dx}{{(x-3)}^2\sqrt{x^2+6x+4}}=\int \frac{1}{{(x-3)}^2\sqrt{{(x-3)}^2-5}}$
Substitute $u=x-3\Rightarrow du=dx$
$I=\int \frac{1}{u^2\sqrt{u^2-5}}du$
Substitute $u=\sqrt{5}\sec t\Rightarrow du=\sqrt{5}\tan t\sec tdt$
$I=\sqrt{5}\int \frac{\cos t}{5\sqrt{5}}dt$
$=\frac{1}{5}\int \cot tdt=\frac{\sin t}{5}$
$=\frac{\sqrt{u^2-5}}{5u}=\frac{\sqrt{x^2-6x+4}}{5(x-3)}$