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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d P}{d t} = 0.2 P (1 - \frac{P}{1000}) - 32

\frac{d}{d x} (lo g_{7} (ln (x)))

\int e^{a r c c o s (x)} d x

\int 2 x - \frac{1}{x ^{2}} d x

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

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