The integral ∫2x−3x2+x+12 dx is equal to - Infinity Learn by Sri Chaitanya

A
$- \frac{1}{x^{2} + x + 1} - \frac{16}{3 \sqrt{3}} \tan^{- 1} \frac{2 x + 1}{\sqrt{3}} - \frac{4}{3} (\frac{2 x + 1}{x^{2} + x + 1}) + C$
B
$- \frac{1}{x^{2} + x + 1} - \frac{16}{3 \sqrt{3}} \tan^{- 1} \frac{2 x - 1}{\sqrt{3}} + \frac{4}{3} (\frac{2 x - 1}{x^{2} + x + 1}) + C$
C
$\frac{1}{2 (x^{2} + x + 1)} - \frac{(2 x + 1)^{2}}{{(x^{2} + x + 1)}^{2}} + C$
D
$\frac{1}{4 (x^{2} + x + 1)} + \frac{2}{3} \tan^{- 1} (2 x + 1) + C$

Solution:

We have,

$\begin{array}{l} I = \int \frac{2 x - 3}{{(x^{2} + x + 1)}^{2}} d x = \int \frac{(2 x + 1) - 4}{{(x^{2} + x + 1)}^{2}} d x \\ \Rightarrow I = \int \frac{2 x + 1}{{(x^{2} + x + 1)}^{2}} d x - 4 \int \frac{1}{{[(x + 1 / 2)^{2} + (\sqrt{3} / 2)^{2}]}^{2}} d x \\ I = - \frac{1}{x^{2} + x + 1} - 4 I_{1} \\ where I_{1} = \int \frac{1}{{[(x + 1 / 2)^{2} + (\sqrt{3} / 2)^{2}]}^{2}} d x \end{array}$

Putting $x + 1 / 2 = (\sqrt{3} / 2) \tan θ in I_{1},$ We get

$\begin{array}{l} I_{1} = \int \frac{\sqrt{3} / 2 \sec^{2} θ d θ}{{[[(\sqrt{3} / 2) \tan θ)]^{2} + (\sqrt{3} / 2)^{2}]}^{2}} \\ \Rightarrow I_{1} = \frac{8}{3 \sqrt{3}} \int \cos^{2} θ d θ = \frac{8}{3 \sqrt{3}} \int \frac{1 + \cos 2 θ}{2} d θ \\ \Rightarrow I_{1} = \frac{4}{3 \sqrt{3}} {θ + \frac{1}{2} \sin 2 θ} + C \\ \Rightarrow I_{1} = \frac{4}{3 \sqrt{3}} \{\tan^{- 1} \frac{2 x + 1}{\sqrt{3}} + \frac{1}{2} \frac{2 \tan θ}{1 + \tan^{2} θ}\} + C \\ \Rightarrow I_{1} = \frac{4}{3 \sqrt{3}} \{\tan^{- 1} \frac{2 x + 1}{\sqrt{3}} + \frac{\sqrt{3}}{4} (\frac{2 x + 1}{x^{2} + x + 1})\} + C \\ ∴ I = - \frac{1}{x^{2} + x + 1} - \frac{16}{3 \sqrt{3}} \tan^{- 1} \frac{2 x + 1}{\sqrt{3}} - \frac{4}{3} (\frac{2 x + 1}{x^{2} + x + 1}) + C \end{array}$

The integral $\int \frac{2 x - 3}{{(x^{2} + x + 1)}^{2}}$ dx is equal to

Solution:

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