Question

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

• A. $2\sqrt{x^2-4x+1}+c$
• B. $log|x-2|+\sqrt{x^2-4x+4}+c$
• C. $\frac{1}{3}\frac{\sqrt{x^2-4x+1}}{x-2}+c$
• D. $si{n}^{-1}\left(\frac{1}{x-2}\right)+c$

Answer 1

The correct option is C $\frac{1}{3}\frac{\sqrt{x^2-4x+1}}{x-2}+c$

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