$\int \frac{2 x - 1}{2 x^{2} + 2 x + 1} d x =$

\frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ + 2 {tan}^{- 1} (2 x + 1) + c

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- \frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ - 2 {tan}^{- 1} (2 x + 1) + c

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- \frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ + 2 {tan}^{- 1} (2 x + 1) + c

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\frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ - 2 {tan}^{- 1} (2 x + 1) + c

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Solution

The correct option is D $\frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ - 2 {tan}^{- 1} (2 x + 1) + c$
$\begin{matrix} \frac{1}{2} [\int \frac{2 x + 1}{x^{2} + x + \frac{1}{2}} d x - \int \frac{2}{x^{2} + x + \frac{1}{2}} d x] x^{2} + x + \frac{1}{2} = t, x + \frac{1}{2} = u = x^{2} + x + \frac{1}{4} = u^{2} (2 x + 1) d x = d t, x + \frac{1}{2} = u \frac{1}{2} [\int \frac{d t}{t} - 2 \int \frac{1}{u^{2} + \frac{1}{u}} d u] = \frac{1}{2} [ln | t | - 2. (\frac{1}{\frac{1}{2}}) {tan}^{- 1} (\frac{u}{\frac{1}{2}})] + c = \frac{1}{2} ln | t | - 2 {tan}^{- 1} 2 u + c = \frac{1}{2} ln ∣ ∣ x^{2} + x + \frac{1}{2} ∣ ∣ - 2 {tan}^{- 1} (2 x + 1) + c = \frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ - 2 {tan}^{- 1} (2 x - 1) + c = \frac{1}{2} ln ∣ ∣ 2 x^{2} + 2 x + 1 ∣ ∣ - 2 {tan}^{- 1} (2 x + 1) + c_{1} c_{1} = c - \frac{1}{2} ln 2 \end{matrix}$

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