Question

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

Answer 1

Answer: 15

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