JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 15)
If $$\int {{{dx} \over {{{({x^2} + x + 1)}^2}}} = a{{\tan }^{ - 1}}\left( {{{2x + 1} \over {\sqrt 3 }}} \right) + b\left( {{{2x + 1} \over {{x^2} + x + 1}}} \right) + C} $$, x > 0 where C is the constant of integration, then the value of $$9\left( {\sqrt 3 a + b} \right)$$ is equal to _____________.
Answer
15
Explanation
$\int \frac{d x}{\left(x^2+x+1\right)^2}$
$=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]^2}$
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2} \tan \theta$
$$ \Rightarrow d x=\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta $$
$\therefore \int \frac{\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta}{\frac{9}{16}\left(\tan ^2 \theta+1\right)^2} $
$ =\frac{8}{3 \sqrt{3}} \int \frac{\sec ^2 \theta d \theta}{\sec ^4 \theta} $
$ =\frac{8}{3 \sqrt{3}} \int \cos ^2 \theta d \theta=\frac{8}{3 \sqrt{3}} \int \frac{1+\cos 2 \theta}{2} d \theta $
$ =\frac{4}{3 \sqrt{3}}\left(\theta+\frac{\sin 2 \theta}{2}\right)+C $
= $ \frac{4}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{4}{3 \sqrt{3}} \frac{\frac{2 x+1}{\sqrt{3}}}{1+\left(\frac{2 x+1}{\sqrt{3}}\right)^2}+C$
$=\frac{4}{3 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{3}\right)+\frac{1}{3} \frac{2 x+1}{\left(x^2+x+1\right)}+C$
$\therefore a=\frac{4}{3 \sqrt{3}}, b=\frac{1}{3}$
Hence, $9(\sqrt{3} a+b)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$
$=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]^2}$
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2} \tan \theta$
$$ \Rightarrow d x=\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta $$
$\therefore \int \frac{\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta}{\frac{9}{16}\left(\tan ^2 \theta+1\right)^2} $
$ =\frac{8}{3 \sqrt{3}} \int \frac{\sec ^2 \theta d \theta}{\sec ^4 \theta} $
$ =\frac{8}{3 \sqrt{3}} \int \cos ^2 \theta d \theta=\frac{8}{3 \sqrt{3}} \int \frac{1+\cos 2 \theta}{2} d \theta $
$ =\frac{4}{3 \sqrt{3}}\left(\theta+\frac{\sin 2 \theta}{2}\right)+C $
= $ \frac{4}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{4}{3 \sqrt{3}} \frac{\frac{2 x+1}{\sqrt{3}}}{1+\left(\frac{2 x+1}{\sqrt{3}}\right)^2}+C$
$=\frac{4}{3 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{3}\right)+\frac{1}{3} \frac{2 x+1}{\left(x^2+x+1\right)}+C$
$\therefore a=\frac{4}{3 \sqrt{3}}, b=\frac{1}{3}$
Hence, $9(\sqrt{3} a+b)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$
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