What is the integral of (6x-9)sqrt(x^2-3x+1) ?

The integral of (6x-9)sqrt(x^2-3x+1) is 2(x^2-3x+1)^{3/2}+C

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integral of (6x-9)sqrt(x^2-3x+1)

\int (6 x - 9) x^{2} - 3 x + 1 d x

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

Solution steps

\int (6 x - 9) x^{2} - 3 x + 1 d x

Apply u-substitution

: \int \frac{u u ^{2} - 45}{36} d u

= \int \frac{u u ^{2} - 45}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u u^{2} - 45 d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{36} \cdot \int v^{2} d v

Apply the Power Rule

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

= \frac{1}{36} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{36} \cdot \frac{( ( 6 x - 9 ) ^{2} - 45 ) ^{3}}{3}

Simplify

\frac{1}{36} \cdot \frac{( ( 6 x - 9 ) ^{2} - 45 ) ^{3}}{3} : 2 (x^{2} - 3 x + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

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What is the integral of (6x-9)sqrt(x^2-3x+1) ?
The integral of (6x-9)sqrt(x^2-3x+1) is 2(x^2-3x+1)^{3/2}+C