Question

The integral $\int \frac{xdx}{2-x^2+\sqrt{2-x^2}}$ equals

• A. $log\left|1+\sqrt{2+x^2}\right|+c$
• B. $-log\left|1+\sqrt{2-x^2}\right|+c$
• C. $-xlog\left|1-\sqrt{2-x^2}\right|+c$
• D. $xlog\left|1-\sqrt{2+x^2}\right|+c$

Answer 1

Answer: (B)

$I=\int \frac{xdx}{2-x^2+\sqrt{2-x^2}}$
Put $t=\sqrt{2-x^2},\frac{dt}{dx}=\frac{1}{2\sqrt{2-x^2}}.\left(-2x\right)$
$\Rightarrow -tdt=xdx$
$\therefore I=\int \frac{\left(-t\right)dt}{t^2+t}=-\int \frac{1}{t+1}dt=-log\left|t+1\right|$
$=-log\left|\sqrt{2-x^2}+1\right|+c$