What is the integral of (z+1)sqrt(3z+2) ?

The integral of (z+1)sqrt(3z+2) is 1/9 (2/5 (3z+2)^{5/2}+2/3 (3z+2)^{3/2})+C

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integral of (z+1)sqrt(3z+2)

\int (z + 1) 3 z + 2 d z

\frac{1}{9} (\frac{2}{5} (3 z + 2)^{\frac{5}{2}} + \frac{2}{3} (3 z + 2)^{\frac{3}{2}}) + C

Solution steps

\int (z + 1) 3 z + 2 d z

Apply u-substitution

: \int \frac{u ( u + 1 )}{9} d u

= \int \frac{u ( u + 1 )}{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 (u + 1) d u

Expand

u (u + 1) : u^{\frac{3}{2}} + u

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

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

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

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 3 z + 2

= \frac{1}{9} (\frac{2}{5} (3 z + 2)^{\frac{5}{2}} + \frac{2}{3} (3 z + 2)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{9} (\frac{2}{5} (3 z + 2)^{\frac{5}{2}} + \frac{2}{3} (3 z + 2)^{\frac{3}{2}}) + C

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What is the integral of (z+1)sqrt(3z+2) ?
The integral of (z+1)sqrt(3z+2) is 1/9 (2/5 (3z+2)^{5/2}+2/3 (3z+2)^{3/2})+C