What is the integral of x(x^2-3)^{7/2} ?

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

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integral of x(x^2-3)^{7/2}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{7}{2}}}{2} d u

= \int \frac{u ^{\frac{7}{2}}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{\frac{7}{2}} d u

Apply the Power Rule

: \frac{2}{9} u^{\frac{9}{2}}

= \frac{1}{2} \cdot \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = x^{2} - 3

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

Simplify

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

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

Add a constant to the solution

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

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