What is the integral of x(3x^2-9)^{5/2} ?

The integral of x(3x^2-9)^{5/2} is 1/21 (3x^2-9)^{7/2}+C

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integral of x(3x^2-9)^{5/2}

\int x (3 x^{2} - 9)^{\frac{5}{2}} d x

\frac{1}{21} (3 x^{2} - 9)^{\frac{7}{2}} + C

Solution steps

\int x (3 x^{2} - 9)^{\frac{5}{2}} d x

Apply u-substitution

= \int \frac{u ^{\frac{5}{2}}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{\frac{5}{2}} d u

Apply the Power Rule

= \frac{1}{6} \cdot \frac{2}{7} u^{\frac{7}{2}}

Substitute back

u = 3 x^{2} - 9

= \frac{1}{6} \cdot \frac{2}{7} (3 x^{2} - 9)^{\frac{7}{2}}

Simplify

\frac{1}{6} \cdot \frac{2}{7} (3 x^{2} - 9)^{\frac{7}{2}} : \frac{1}{21} (3 x^{2} - 9)^{\frac{7}{2}}

= \frac{1}{21} (3 x^{2} - 9)^{\frac{7}{2}}

Add a constant to the solution

= \frac{1}{21} (3 x^{2} - 9)^{\frac{7}{2}} + C

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What is the integral of x(3x^2-9)^{5/2} ?
The integral of x(3x^2-9)^{5/2} is 1/21 (3x^2-9)^{7/2}+C