What is the integral of sqrt((x-7)^3) ?

The integral of sqrt((x-7)^3) is 2/5 (x-7)^{5/2}+C

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integral of sqrt((x-7)^3)

\int (x - 7)^{3} d x

\frac{2}{5} (x - 7)^{\frac{5}{2}} + C

Solution steps

\int (x - 7)^{3} d x

(x - 7)^{3} = (x - 7)^{\frac{3}{2}},

assuming

(x - 7) \geq 0

= \int (x - 7)^{\frac{3}{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x - 7

= \frac{2}{5} (x - 7)^{\frac{5}{2}}

Add a constant to the solution

= \frac{2}{5} (x - 7)^{\frac{5}{2}} + C

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What is the integral of sqrt((x-7)^3) ?
The integral of sqrt((x-7)^3) is 2/5 (x-7)^{5/2}+C