What is the integral of (x^2-x)/(x^3-5x^2+3x+9) ?

The integral of (x^2-x)/(x^3-5x^2+3x+9) is 1/8 ln|x+1|+7/8 ln|x-3|-3/(2(x-3))+C

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integral of (x^2-x)/(x^3-5x^2+3x+9)

\int \frac{x ^{2} - x}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

\frac{1}{8} ln ∣ x + 1 ∣ + \frac{7}{8} ln ∣ x - 3 ∣ - \frac{3}{2 ( x - 3 )} + C

Solution steps

\int \frac{x ^{2} - x}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

Take the partial fraction of

\frac{x ^{2} - x}{x ^{3} - 5 x ^{2} + 3 x + 9} : \frac{1}{8 ( x + 1 )} + \frac{7}{8 ( x - 3 )} + \frac{3}{2 ( x - 3 ) ^{2}}

= \int \frac{1}{8 ( x + 1 )} + \frac{7}{8 ( x - 3 )} + \frac{3}{2 ( x - 3 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{8 ( x + 1 )} d x + \int \frac{7}{8 ( x - 3 )} d x + \int \frac{3}{2 ( x - 3 ) ^{2}} d x

\int \frac{1}{8 ( x + 1 )} d x = \frac{1}{8} ln ∣ x + 1 ∣

\int \frac{7}{8 ( x - 3 )} d x = \frac{7}{8} ln ∣ x - 3 ∣

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

= \frac{1}{8} ln ∣ x + 1 ∣ + \frac{7}{8} ln ∣ x - 3 ∣ - \frac{3}{2 ( x - 3 )}

Add a constant to the solution

= \frac{1}{8} ln ∣ x + 1 ∣ + \frac{7}{8} ln ∣ x - 3 ∣ - \frac{3}{2 ( x - 3 )} + C

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What is the integral of (x^2-x)/(x^3-5x^2+3x+9) ?
The integral of (x^2-x)/(x^3-5x^2+3x+9) is 1/8 ln|x+1|+7/8 ln|x-3|-3/(2(x-3))+C