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If $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = a t a n^{- 1} (\frac{2 x + 1}{\sqrt[]{3}}) + b (\frac{2 x + 1}{x^{2} + x + 1}) + C, x > 0$ where C is the constant of integration, then the value of 9 $(\sqrt[]{3} a + b)$ is equal to

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Vishal Baghel | Contributor-Level 10

a month ago

Let $l = \int \frac{d x}{{(x^{2} + x + 1)}^{2}} . . . . . . . . . . . . . (i)$

Now $\int \frac{d x}{x^{2} + x + 1} = \int (\frac{1}{x^{2} + x + 1}) d x$

by integration by parts

=> $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = \frac{4}{3 \sqrt[]{2}} t a n^{- 1} \frac{2 x + 1}{\sqrt[]{3}} + \frac{1}{3} \frac{(2 x + 1)}{(x^{2} + x + 1)} + c$

$\Rightarrow a = \frac{4}{3 \sqrt[]{3}}, b = \frac{1}{3}$

Now, $9 (\sqrt[]{3} a + b) = 9 (\frac{4}{3} + \frac{1}{3}) = 15$

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If $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = a t a n^{- 1} (\frac{2 x + 1}{\sqrt[]{3}}) + b (\frac{2 x + 1}{x^{2} + x + 1}) + C, x > 0$ where C is the constant of integration, then the value of 9 $(\sqrt[]{3} a + b)$ is equal to

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If ∫ d x ( x 2 + x + 1 ) 2 = a t a n − 1 ( 2 x + 1 3 ) + b ( 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

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If $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = a t a n^{- 1} (\frac{2 x + 1}{\sqrt[]{3}}) + b (\frac{2 x + 1}{x^{2} + x + 1}) + C, x > 0$ where C is the constant of integration, then the value of 9 $(\sqrt[]{3} a + b)$ is equal to