What is the integral of (x^2-4)/(x^3-3x^2-x+3) ?

The integral of (x^2-4)/(x^3-3x^2-x+3) is 5/8 ln|x-3|-3/8 ln|x+1|+3/4 ln|x-1|+C

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integral of (x^2-4)/(x^3-3x^2-x+3)

\int \frac{x ^{2} - 4}{x ^{3} - 3 x ^{2} - x + 3} d x

\frac{5}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{2} - 4}{x ^{3} - 3 x ^{2} - x + 3} d x

Take the partial fraction of

\frac{x ^{2} - 4}{x ^{3} - 3 x ^{2} - x + 3} : \frac{5}{8 ( x - 3 )} - \frac{3}{8 ( x + 1 )} + \frac{3}{4 ( x - 1 )}

= \int \frac{5}{8 ( x - 3 )} - \frac{3}{8 ( x + 1 )} + \frac{3}{4 ( x - 1 )} 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{5}{8 ( x - 3 )} d x - \int \frac{3}{8 ( x + 1 )} d x + \int \frac{3}{4 ( x - 1 )} d x

\int \frac{5}{8 ( x - 3 )} d x = \frac{5}{8} ln ∣ x - 3 ∣

\int \frac{3}{8 ( x + 1 )} d x = \frac{3}{8} ln ∣ x + 1 ∣

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

= \frac{5}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{5}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + C

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

(4 - t) (4 + t^{3})^{- 1}

\frac{\partial}{\partial y} (\frac{4}{3} π (x^{3} - y^{3}))

\frac{d}{d x} (\frac{8 + csc ( x )}{7 - csc ( x )})

x \to \infty lim (\frac{3}{5 x})

What is the integral of (x^2-4)/(x^3-3x^2-x+3) ?
The integral of (x^2-4)/(x^3-3x^2-x+3) is 5/8 ln|x-3|-3/8 ln|x+1|+3/4 ln|x-1|+C