Question

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

Answer 1

\[Let I = \int\left( x + 2 \right) \sqrt{3x + 5}  \text{dx} \]

\[\text{Putting 3x + 5 }= t\]
\[ \Rightarrow x = \frac{t - 5}{3}\]

\[\Rightarrow 3dx = dt\]
\[ \Rightarrow dx = \frac{dt}{3}\]

` ∴ I = ∫ ( {t-5} /3 +2) \sqrt t    dt/3 `
`  =1/3   ∫ ( {t-5+6} /3 ) \sqrt t    dt `
\[ = \frac{1}{9}\int\left( t^\frac{3}{2} + t^\frac{1}{2} \right) dt\]
\[ = \frac{1}{9}\left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} + \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{9}\left[ \frac{2}{5} t^\frac{5}{2} + \frac{2}{3} t^\frac{3}{2} \right] + C\]
\[ = \frac{1}{9}\left[ \frac{2}{5} \left( 3x + 5 \right)^\frac{5}{2} + \frac{2}{3} \left( 3x + 5 \right)^\frac{3}{2} \right] + C \left[ \because t = 3x + 5 \right]\]
\[ = \frac{2}{9}\left[ \left( 3x + 5 \right)^\frac{3}{2} \left\{ \frac{3x + 5}{5} + \frac{1}{3} \right\} \right] + C\]
\[ = \frac{2}{9}\left[ \left( 3x + 5 \right)^\frac{3}{2} \left\{ \frac{9x + 15 + 5}{15} \right\} \right] + C\]
\[ = \frac{2}{9}\left[ \left( 3x + 5 \right)^\frac{3}{2} \left\{ \frac{9x + 20}{15} \right\} \right] + C\]
\[ = \frac{2}{135} \left( 3x + 5 \right)^\frac{3}{2} \left( 9x + 20 \right) + C\]