What is the integral of (x^3+3x^2-18x)/((x-3)(x+6)) ?

The integral of (x^3+3x^2-18x)/((x-3)(x+6)) is (x^2)/2+C

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integral of (x^3+3x^2-18x)/((x-3)(x+6))

\int \frac{x ^{3} + 3 x ^{2} - 18 x}{( x - 3 ) ( x + 6 )} d x

\frac{x ^{2}}{2} + C

Solution steps

\int \frac{x ^{3} + 3 x ^{2} - 18 x}{( x - 3 ) ( x + 6 )} d x

Simplify

\frac{x ^{3} + 3 x ^{2} - 18 x}{( x - 3 ) ( x + 6 )} : x

= \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{2}}{2} + C

\int \frac{x + 2}{2} d x

\int \frac{20 ( 1 - x ^{2} )}{1 + x ^{2}} d x

\int \frac{( x )}{x} d x

\int sin^{6} (\frac{3 x}{2}) d x

\int (\frac{8}{3}) d m

What is the integral of (x^3+3x^2-18x)/((x-3)(x+6)) ?
The integral of (x^3+3x^2-18x)/((x-3)(x+6)) is (x^2)/2+C