What is the integral of (x^2+4x-5)/((x^2-9)(x+1)) ?

The integral of (x^2+4x-5)/((x^2-9)(x+1)) is -2/3 ln|x+3|+2/3 ln|x-3|+ln|x+1|+C

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integral of (x^2+4x-5)/((x^2-9)(x+1))

\int \frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} d x

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

Solution steps

\int \frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} d x

Take the partial fraction of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} : - \frac{2}{3 ( x + 3 )} + \frac{2}{3 ( x - 3 )} + \frac{1}{x + 1}

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

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

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

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

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

Add a constant to the solution

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

What is the integral of (x^2+4x-5)/((x^2-9)(x+1)) ?
The integral of (x^2+4x-5)/((x^2-9)(x+1)) is -2/3 ln|x+3|+2/3 ln|x-3|+ln|x+1|+C