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Question
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
Sum
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Solution
\[\int\left( 4x + 2 \right) \sqrt{x^2 + x + 1} \text{dx}\]
\[ = 2\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} dx\]
\[\text{Let }x^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]
\[Now, 2\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} dx\]
\[ = 2\int\sqrt{t} \text{dt}\]
\[ = 2\int t^\frac{1}{2} \text{dt}\]
\[ = 2 \left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = 2 \times \frac{2}{3} t^\frac{3}{2} + C\]
\[ = \frac{4}{3} \text{t}^\frac{3}{2} + C\]
\[ = \frac{4}{3} \left( x^2 + x + 1 \right)^\frac{3}{2} + C\]
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