Question

If $\int \frac{dx}{(x^2+x+1{)}^2}=a{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+b\left(\frac{2x+1}{x^2+x+1}\right)+C$, $x>0$ where $C$ is the constant of integration, then the value of $9(\sqrt{3}a+b)$ is equal to

Answer 1

$\int \frac{dx}{(x^2+x+1{)}^2}$
$=\int \frac{dx}{{({(x+\frac{1}{2})}^2+\frac{3}{4})}^2}$
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan \theta$
$\Rightarrow dx=\frac{\sqrt{3}}{2}{\sec }^2\theta$

$\therefore \int \frac{\frac{\sqrt{3}}{2}.{\sec }^2\theta }{\frac{9}{16}{\sec }^4\theta }d\theta$
$=\frac{8}{3\sqrt{3}}\int {\cos }^2\theta d\theta$
$=\frac{4}{3\sqrt{3}}\int 2{\cos }^2\theta d\theta$
$=\frac{4}{3\sqrt{3}}\int (1+\cos 2\theta )d\theta$
$=\frac{4}{3\sqrt{3}}(\theta +\frac{\sin 2\theta }{2})+C$
Now, $\tan \theta =\frac{2x+1}{\sqrt{3}}$
$\Rightarrow \sin 2\theta =2\left(\sin \theta \right)\left(\cos \theta \right)$
$\Rightarrow \sin 2\theta =\frac{\sqrt{3}(2x+1)}{2(x^2+x+1)}$

$\therefore \frac{4}{3\sqrt{3}}({\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{\sqrt{3}}{4}\left(\frac{2x+1}{x^2+x+1}\right))+C$
$=\frac{4}{3\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{1}{3}\left(\frac{2x+1}{x^2+x+1}\right)+C$
$\Rightarrow a=\frac{4}{3\sqrt{3}},b=\frac{1}{3}$
$\therefore 9(\sqrt{3}a+b)=15$