Question

\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]

Answer 1

\[\int\left( \frac{2x + 1}{\sqrt{3x + 2}} \right)dx\]
\[ = \frac{1}{3}\int\left( \frac{6x + 3}{\sqrt{3x + 2}} \right)dx\]
\[ = \frac{1}{3}\int\left( \frac{6x + 4 - 1}{\sqrt{3x + 2}} \right)dx\]
\[ = \frac{1}{3}\int\left( \frac{2\left( 3x + 2 \right)}{\sqrt{3x + 2}} - \frac{1}{\sqrt{3x + 2}} \right)dx\]
\[ = \frac{1}{3}\int\left( 2\sqrt{3x + 2} - \frac{1}{\sqrt{3x + 2}} \right)dx\]
\[ = \frac{1}{3}\left[ \int2 \left( 3x + 2 \right)^\frac{1}{2} dx - \int \left( 3x + 2 \right)^{- \frac{1}{2}} dx \right]\]
\[ = \frac{1}{3}\left[ 2\left\{ \frac{\left( 3x + 2 \right)^\frac{1}{2} + 1}{3 \left( \frac{1}{2} + 1 \right)} \right\} - \frac{\left( 3x + 2 \right)^{- \frac{1}{2} + 1}}{\left( - \frac{1}{2} + 1 \right) \times 3} \right] + C\]
\[ = \frac{1}{3}\left[ \frac{4}{9} \left( 3x + 2 \right)^\frac{3}{2} - \frac{2}{3} \left( 3x + 2 \right)^\frac{1}{2} \right] + C\]
\[ = \frac{4}{27} \left( 3x + 2 \right)^\frac{3}{2} - \frac{2}{9} \left( 3x + 2 \right)^\frac{1}{2} + C\]
\[ = \sqrt{3x + 2}\left( \frac{4}{27}\left( 3x + 2 \right) - \frac{2}{9} \right) + C\]
\[ = \sqrt{3x + 2}\left( \frac{4\left( 3x + 2 \right) - 6}{27} \right) + C\]
\[ = \sqrt{3x + 2}\left( \frac{12x + 8 - 6}{27} \right) + C\]
\[ = \frac{2}{27}\left( 6x + 1 \right)\sqrt{3x + 2} + C\]