Question

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

Answer 1

\[\int\left( \frac{x + 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 2}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3 - 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3}{\sqrt{2x + 3}} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \sqrt{2x + 3} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\left[ \int \left( 2x + 3 \right)^\frac{1}{2} dx - \int \left( 2x + 3 \right)^{- \frac{1}{2}} dx \right]\]
\[ = \frac{1}{2}\left[ \frac{\left( 2x + 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} - \frac{\left( 2x + 3 \right)^{- \frac{1}{2} + 1}}{2\left( - \frac{1}{2} + 1 \right)} + C \right]\]
\[ = \frac{1}{2}\left[ \frac{1}{3} \left( 2x + 3 \right)^\frac{3}{2} - \left( 2x + 3 \right)^\frac{1}{2} + C \right]\]
\[ = \frac{1}{6} \left( 2x + 3 \right)^\frac{3}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{1}{2} + C\]