Question

Integrate : (3x+4) √3x+7 dx​

Answer 1

Step-by-step explanation:

$\displaystyle{\int {(3x + 4) \sqrt{(3x + 7)} dx}} \\ let \: \\ 3x + 7 = t \\ 3x = t - 7 \\ x = \dfrac{1}{3} (t - 7) \\ dx = \dfrac{1}{3} (dt - 0) \\ dx = \dfrac{1}{3} dt \\ now \\ \displaystyle \int(t - 7 + 4) \: \sqrt{t} \: \: \frac{dt}{3} \\ = \dfrac{1}{3} \displaystyle \int (t - 3) \: \: {t}^{ \frac{1}{2} } \: dt \\ = \dfrac{1}{3} \displaystyle \int( {t}^{ \frac{3}{2} } - 3 {t}^{ \frac{1}{2} } )dt \\ = \dfrac{1}{3} ( \: \dfrac{ {t}^{ \frac{3}{2} + 1} }{ \frac{3}{2} + 1} - 3 \dfrac{ {t}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1} ) + c \\ = \dfrac{1}{3} ( \: \frac{ {t}^{ \frac{5}{2} } }{ \frac{5}{2} } - 3 \dfrac{ {t}^{ \frac{3}{2} } }{ \frac{3}{2} } ) + c \\ = \dfrac{2}{15} {t}^{ \frac{5}{2} } - \dfrac{2}{3} {t}^{ \frac{3}{2} } + c$