QJEEM21Ma76b7: JEE Main 2021 Mathematics - Class 12 JEEM21Ma76b7 Solution: If ∫ d x ( x 2 + x + 1 ) 2 = a tan − 1 ( 2 x + 1 3 ) + b ( 2 x…

Question

Verified Solution • 2025–26 Session moderate

If

\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = a \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + b (\frac{2 x + 1}{x^{2} + x + 1}) + C

, x > 0 where C is the constant of integration, then the value of

9 (\sqrt{3} a + b)

is equal to _____________.

Accepted Answer

\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} θ d θ

\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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