What is the integral of (x^{3/2}+6)^8sqrt(x) ?

The integral of (x^{3/2}+6)^8sqrt(x) is 2/27 (x^{3/2}+6)^9+C

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integral of (x^{3/2}+6)^8sqrt(x)

\int (x^{\frac{3}{2}} + 6)^{8} x d x

\frac{2}{27} (x^{\frac{3}{2}} + 6)^{9} + C

Solution steps

\int (x^{\frac{3}{2}} + 6)^{8} x d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{2}{3} \cdot \frac{u ^{9}}{9}

Substitute back

u = x^{\frac{3}{2}} + 6

= \frac{2}{3} \cdot \frac{( x ^{\frac{3}{2}} + 6 ) ^{9}}{9}

Simplify

\frac{2}{3} \cdot \frac{( x ^{\frac{3}{2}} + 6 ) ^{9}}{9} : \frac{2}{27} (x^{\frac{3}{2}} + 6)^{9}

= \frac{2}{27} (x^{\frac{3}{2}} + 6)^{9}

Add a constant to the solution

= \frac{2}{27} (x^{\frac{3}{2}} + 6)^{9} + C

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What is the integral of (x^{3/2}+6)^8sqrt(x) ?
The integral of (x^{3/2}+6)^8sqrt(x) is 2/27 (x^{3/2}+6)^9+C