What is the integral of (5x-7)/((x^2-x-2)(x-3)) ?

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

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integral of (5x-7)/((x^2-x-2)(x-3))

\int \frac{5 x - 7}{( x ^{2} - x - 2 ) ( x - 3 )} d x

- ln ∣ x + 1 ∣ - ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{5 x - 7}{( x ^{2} - x - 2 ) ( x - 3 )} d x

Take the partial fraction of

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

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

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

= - ln ∣ x + 1 ∣ - ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ - ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣ + C

\frac{d}{d x} (6 ln (x) + 11)

\int_{- 0.5}^{1} cos (x^{2}) d x

\frac{d y}{d x} = \frac{y}{x} + 6 x + 7

x \to 4 lim (\frac{2 x + 8}{x - 4})

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

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