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

The integral of (2x^2-6)/(sqrt(6+9x-x^3)) is -4/3 sqrt(6+9x-x^3)+C

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

\int \frac{2 x ^{2} - 6}{6 + 9 x - x ^{3}} d x

- \frac{4}{3} 6 + 9 x - x^{3} + C

Solution steps

\int \frac{2 x ^{2} - 6}{6 + 9 x - x ^{3}} d x

Apply u-substitution

: \int - \frac{2}{3 u} d u

= \int - \frac{2}{3 u} 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 \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - \frac{2}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 + 9 x - x^{3}

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

Simplify

- \frac{2}{3} \cdot 2 (6 + 9 x - x^{3})^{\frac{1}{2}} : - \frac{4}{3} 6 + 9 x - x^{3}

= - \frac{4}{3} 6 + 9 x - x^{3}

Add a constant to the solution

= - \frac{4}{3} 6 + 9 x - x^{3} + C

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What is the integral of (2x^2-6)/(sqrt(6+9x-x^3)) ?
The integral of (2x^2-6)/(sqrt(6+9x-x^3)) is -4/3 sqrt(6+9x-x^3)+C