What is the integral of (4x-6)sqrt(x^2-3x) ?

The integral of (4x-6)sqrt(x^2-3x) is 4/3 (x(x-3))^{3/2}+C

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integral of (4x-6)sqrt(x^2-3x)

\int (4 x - 6) x^{2} - 3 x d x

\frac{4}{3} (x (x - 3))^{\frac{3}{2}} + C

Solution steps

\int (4 x - 6) x^{2} - 3 x d x

Apply u-substitution

: \int \frac{u u ^{2} - 36}{16} d u

= \int \frac{u u ^{2} - 36}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u u^{2} - 36 d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{16} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{16} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{16} \cdot \frac{( ( 4 x - 6 ) ^{2} - 36 ) ^{3}}{3}

Simplify

\frac{1}{16} \cdot \frac{( ( 4 x - 6 ) ^{2} - 36 ) ^{3}}{3} : \frac{4}{3} (x (x - 3))^{\frac{3}{2}}

= \frac{4}{3} (x (x - 3))^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{3} (x (x - 3))^{\frac{3}{2}} + C

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What is the integral of (4x-6)sqrt(x^2-3x) ?
The integral of (4x-6)sqrt(x^2-3x) is 4/3 (x(x-3))^{3/2}+C