Question

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

Answer 1

\[\text{Let I} = \int\left( 5x + 3 \right) \sqrt{2x - 1}dx\]
\[Putting\ 2x - 1 = t\]
\[ \Rightarrow 2x = t + 1\]
\[ \Rightarrow x = \frac{t + 1}{2}\]

\[\text{and}\ 2dx = dt\]
\[ \Rightarrow dx = \frac{dt}{2}\]

\[\therefore I = \int\left[ 5\left( \frac{t + 1}{2} \right) + 3 \right] \cdot \sqrt{t} \cdot \frac{dt}{2}\]

` = ∫ ( 5t / 2 + 5/2 + 3 ) ×( \sqrt t   dt) /2 `
\[ = \frac{1}{4}\int\left( 5t + 11 \right) t^\frac{1}{2} dt\]
\[ = \frac{1}{4}\int\left( 5 t^\frac{3}{2} + 11 t^\frac{1}{2} \right) dt\]
\[ = \frac{1}{4}\left[ 5\frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} + 11\frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]

\[ = \frac{1}{4} \times \frac{2}{5} \times  \text{5 } t^\frac{5}{2} + \frac{1}{4} \times 11 \times \frac{2}{3}\text{ t}^\frac{3}{2} + C\]
\[ = \frac{1}{2} t^\frac{5}{2} + \frac{11}{6} t^\frac{3}{2} + C\]

\[ = \frac{t^\frac{3}{2}}{2}\left[ t + \frac{11}{3} \right] + C\]

\[ = \frac{t^\frac{3}{2}}{2}\left[ \frac{3t + 11}{3} \right] + C\]

\[ = \frac{\left( 2x - 1 \right)^\frac{3}{2}}{2}\left[ \frac{3\left( 2x - 1 \right) + 11}{3} \right] + C \left[ \because t = 2x - 1 \right]\]

\[ = \frac{\left( 2x - 1 \right)^\frac{3}{2}}{2}\left[ \frac{6x - 3 + 11}{3} \right] + C\]

\[ = \left( \frac{2x - 1}{2} \right)^\frac{3}{2} \left[ \frac{2 \left( 3x + 4 \right)}{3} \right] + C\]

\[ = \frac{\left( 2x - 1 \right)^\frac{3}{2} \left( 3x + 4 \right)}{3} + C\]