Question

${\int }_0^1\frac{x(2x^2+1)}{x^8+2x^6-x^2+1}dx=$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2\sqrt{3}}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{2\sqrt{3}}$

Let $I={\int }_0^1\frac{x(2x^2+1)}{x^8+2x^6-x^2+1}dx$

$={\int }_0^1\frac{(2x^3+x)}{(x^4+x^2{)}^2-(x^4+x^2)+1}dx$
Put $t=x^4+x^2$
$\Rightarrow dt=4x^3+2x$
Also $t=0$ for $x=0$ and $t=2$ for $x=1$
So, $I=\frac{1}{2}{\int }_0^2\frac{dt}{t^2-t+1}$
$=\frac{1}{2}{\int }_0^2\frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=\frac{1}{2}.\frac{2}{\sqrt{3}}{\left\{ta{n}^{-1}\frac{2(t-\frac{1}{2})}{\sqrt{3}}\right\}}_0^2$
$=\frac{1}{\sqrt{3}}\left(ta{n}^{-1}\right(\sqrt{3})-ta{n}^{-1}(-\frac{1}{\sqrt{3}}))$
$=\frac{1}{\sqrt{3}}\left(\frac{\pi }{3}+\frac{\pi }{6}\right)$
$=\frac{\pi }{2\sqrt{3}}$