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JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Indefinite Integrals

If

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

is equal to

Answer: 15

Solution

With

X = x + \frac{1}{2}

and

k = \frac{3}{2}

,

x^{2} + x + 1 = X^{2} + k^{2}

and

\int \frac{d X}{( X ^{2} + k ^{2} ) ^{2}} = \frac{X}{2 k ^{2} ( X ^{2} + k ^{2} )} + \frac{1}{2 k ^{3}} tan^{- 1} \frac{X}{k} + C .

Here

\frac{X}{2 k ^{2} ( X ^{2} + k ^{2} )} = \frac{2 x + 1}{3 ( x ^{2} + x + 1 )}

and

\frac{1}{2 k ^{3}} = \frac{4}{3 3}

with

\frac{X}{k} = \frac{2 x + 1}{3}

. Comparing,

a = \frac{4}{3 3}

and

b = \frac{1}{3}

, so

9 (3 a + b) = 9 (\frac{4}{3} + \frac{1}{3}) = 15.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer

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