What is the integral of t(3t+2)^{3/2} ?

The integral of t(3t+2)^{3/2} is 1/9 (2/7 (3t+2)^{7/2}-4/5 (3t+2)^{5/2})+C

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integral of t(3t+2)^{3/2}

\int t (3 t + 2)^{\frac{3}{2}} d t

\frac{1}{9} (\frac{2}{7} (3 t + 2)^{\frac{7}{2}} - \frac{4}{5} (3 t + 2)^{\frac{5}{2}}) + C

Solution steps

\int t (3 t + 2)^{\frac{3}{2}} d t

Apply u-substitution

: \int \frac{u ^{\frac{3}{2}} ( u - 2 )}{9} d u

= \int \frac{u ^{\frac{3}{2}} ( u - 2 )}{9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int u^{\frac{3}{2}} (u - 2) d u

Expand

u^{\frac{3}{2}} (u - 2) : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}}

= \frac{1}{9} \cdot \int u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (\int u^{\frac{5}{2}} d u - \int 2 u^{\frac{3}{2}} d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

= \frac{1}{9} (\frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}})

Substitute back

u = 3 t + 2

= \frac{1}{9} (\frac{2}{7} (3 t + 2)^{\frac{7}{2}} - \frac{4}{5} (3 t + 2)^{\frac{5}{2}})

Add a constant to the solution

= \frac{1}{9} (\frac{2}{7} (3 t + 2)^{\frac{7}{2}} - \frac{4}{5} (3 t + 2)^{\frac{5}{2}}) + C

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What is the integral of t(3t+2)^{3/2} ?
The integral of t(3t+2)^{3/2} is 1/9 (2/7 (3t+2)^{7/2}-4/5 (3t+2)^{5/2})+C