If ∫ dx ( x ^ 2 + x + 1) ^ 2 = a ^ - 1 ( 2x + 1 3 ) + b( 2x + 1 x ^ 2 + x + 1 ) + C, x > 0 where C is the constant of integration, then the value of 9 ( 3 a + b ) is equal to _________.

Correct answer: 0 I = ∫ dx [ (x + (1)/(2)) ^ 2 + (3)/(4)] ^ 2 Put x + (1)/(2) = t = ∫ dt ( t ^ 2 + (3)/(4)) ^ 2 [Putt = 3 2 tanθ] = 3 2 ∫ ^ 2 θ (9)/(16) ^ 4 θ = 4 3 9 ∫ (1 + 2θ)dθ = 4 3 9 [θ + ( 2θ)/(2)] + C = 4 3 9 [ ^ - 1 ((2x + 1)/(3)) + 3 (2x + 1) 3 + (2x + 1) ^ 2 ] + C = 4 3 9 ^ - 1 ( 2x + 1 3 ) + (1)/(3)( 2x + 1 x ^ 2 + x + 1 ) + C Hence, 9( 3 a + b) = 15

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Indefinite Integration

If ∫ dx ( x ^ 2 + x + 1) ^ 2 = a ^ - 1 ( 2x + 1 3 ) + b( 2x + 1 x ^...

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

Correct answer

Step-by-step solution

I = ∫ \dfrac{dx}{{\left[{\left(x + \dfrac{1}{2}\right)}^{2} + \dfrac{3}{4}\right]}^{2}} $Put$ x + \dfrac{1}{2} = t $$ = ∫ \dfrac{dt}{{\left({t}^{2} + \dfrac{3}{4}\right)}^{2}} $\left[Putt = \dfrac{\sqrt{3}}{2}tan\theta\right] = \dfrac{\sqrt{3}}{2} ∫ \dfrac{\sec^{2}\thetad\theta}{\dfrac{9}{16}\sec^{4}\theta}$

$= \dfrac{4\sqrt{3}}{9} ∫ \left(1 + \cos 2\theta\right)d\theta$

$= \dfrac{4\sqrt{3}}{9}\left[\theta + \dfrac{\sin 2\theta}{2}\right] + C$

$= \dfrac{4\sqrt{3}}{9}\left[\tan^{- 1}\left(\dfrac{2x + 1}{3}\right) + \dfrac{\sqrt{3}\left(2x + 1\right)}{3 + {\left(2x + 1\right)}^{2}}\right] + C$

$= \dfrac{4\sqrt{3}}{9}\tan^{- 1}\left(\dfrac{2x + 1}{\sqrt{3}}\right) + \dfrac{1}{3}\left(\dfrac{2x + 1}{{x}^{2} + x + 1}\right) + C$

Hence, $9\left(\sqrt{3}a + b\right) = 15$

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