Question

\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\left( \frac{8x + 13}{\sqrt{4x + 7}} \right)dx\]
\[\text{ Putting  4x} + 7 = t\]
\[ \Rightarrow x = \frac{t - 7}{4}\]
\[ \Rightarrow 4dx = dt\]
\[ \Rightarrow dx = \frac{dt}{4}\]
\[ \therefore I = \frac{1}{4}\int\left\{ \frac{8 \left( \frac{t - 7}{4} \right) + 13}{\sqrt{t}} \right\}dt\]
\[ = \frac{1}{4}\int\left( \frac{2t - 14 + 13}{\sqrt{t}} \right)dt\]
\[ = \frac{1}{4}\int\left( \frac{2t - 1}{\sqrt{t}} \right)dt\]
\[ = \frac{1}{4}\int\frac{2t}{\sqrt{t}}dt - \frac{1}{4}\int\frac{dt}{\sqrt{t}}\]
\[ = \frac{1}{2}\int t^\frac{1}{2} dt - \frac{1}{4}\int t^{- \frac{1}{2}} dt\]
\[ = \frac{1}{2}\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] - \frac{1}{4}\left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{2} \times \frac{2}{3} t^\frac{3}{2} - \frac{2}{4} t^\frac{1}{2} + C\]
\[ = \frac{1}{3} t^\frac{3}{2} - \frac{2}{4} t^\frac{1}{2} + C\]
\[ = \frac{1}{3} \left( 4x + 7 \right)^\frac{3}{2} - \frac{1}{2} \left( 4x + 7 \right)^\frac{1}{2} + C ....................\left( \because t = 4x + 7 \right)\]
\[ = \frac{1}{3} \left( 4x + 7 \right)^\frac{3}{2} - \sqrt{4x + 7} + C\]