Question

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

Answer 1

$I =\int \frac{ dx }{\left[\left( x +\frac{1}{2}\right)^{2}+\frac{3}{4}\right]^{2}}$
$\int \frac{ dt }{\left( t ^{2}+\frac{3}{4}\right)^{2}}\left(\right.$ Put $\left.x+\frac{1}{2}=t\right)$
$=\frac{\sqrt{3}}{2} \int \frac{\sec ^{2} \theta\, d\,\theta}{\frac{9}{16} \sec ^{4} \theta}\left(\right.$ Put $\left.t =\frac{\sqrt{3}}{2} \tan \theta\right)$
$=\frac{4 \sqrt{3}}{9} \int(1+\cos 2 \,\theta) d \,\theta$
$=\frac{4 \sqrt{3}}{9}\left[\theta+\frac{\sin 2\, \theta}{2}\right]+c$
$=\frac{4 \sqrt{3}}{9}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{\sqrt{3}(2 x+1)}{3+(2 x+1)^{2}}\right]+c$
$=\frac{4 \sqrt{3}}{9} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{1}{3}\left(\frac{2 x+1}{x^{2}+x+1}\right)+c$
Hence, $9(\sqrt{3} a+b)=15$