Question

${\int }_{-1}^1\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}dx$

• A. $\frac{3\pi }{2}$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

$f\left(x\right)=\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$
$f(-x)=\frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}}=-f\left(x\right)$
Since, $f\left(x\right)=-f(-x)$ this is an odd function so answer will be $0$.