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

y^{^{'}} + x y = - x^{2}

y = (5 x - 4)^{5} (3 - x^{3})^{5}

f (x) = 2 x^{2} - 6 x, at x = 1

g (x) = 5 x + 20, at x = 1

4 x + 3

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