What is the integral of (x-5)/(x^3-3x^2+4) ?

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

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integral of (x-5)/(x^3-3x^2+4)

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 4} d x

- \frac{2}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{x - 2} + C

Solution steps

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 4} d x

Take the partial fraction of

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

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

\int \frac{2}{3 ( x + 1 )} d x = \frac{2}{3} ln ∣ x + 1 ∣

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

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

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

Simplify

= - \frac{2}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{x - 2}

Add a constant to the solution

= - \frac{2}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{x - 2} + C

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