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