Q107 / 160 · AP EAMCET 2015 · Mathematics
$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$ is equal to
- A$\frac{3\pi}{2}$
- B$\frac{\pi}{2}$
- Correct answerC$0$
- D$-1$
Solution
$I=\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$
Let $f(x)=\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$
Replacing $x$ by $-x$, we get
$\begin{aligned}
& f(-x)=\frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}} \\
& =\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}=-f(x)
\end{aligned}$
So, $f(x)$ is an odd function.
$\therefore \quad \int_{-1}^1 f(x) d x=0$
Hence, the correct answer is option: 'C'.