What is the integral of (8x+4)^{3/2} ?

The integral of (8x+4)^{3/2} is 8/5 (2x+1)^{5/2}+C

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integral of (8x+4)^{3/2}

\int (8 x + 4)^{\frac{3}{2}} d x

\frac{8}{5} (2 x + 1)^{\frac{5}{2}} + C

Solution steps

\int (8 x + 4)^{\frac{3}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{2}}}{8} d u

= \int \frac{u ^{\frac{3}{2}}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{\frac{3}{2}} d u

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= \frac{1}{8} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = 8 x + 4

= \frac{1}{8} \cdot \frac{2}{5} (8 x + 4)^{\frac{5}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{5} (8 x + 4)^{\frac{5}{2}} : \frac{8}{5} (2 x + 1)^{\frac{5}{2}}

= \frac{8}{5} (2 x + 1)^{\frac{5}{2}}

Add a constant to the solution

= \frac{8}{5} (2 x + 1)^{\frac{5}{2}} + C

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What is the integral of (8x+4)^{3/2} ?
The integral of (8x+4)^{3/2} is 8/5 (2x+1)^{5/2}+C