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. $-x\,log\left|1-\sqrt{2-x^{2}}\right|+c$
• D. $x\,log\left|1-\sqrt{2+x^{2}}\right|+c$

Answer 1

The Correct Option is B

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