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

The integral of (x^2+2x-4)/((x+3)(x+4)(x-5)) is 1/8 ln|x+3|+4/9 ln|x+4|+31/72 ln|x-5|+C

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

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

\frac{1}{8} ln ∣ x + 3 ∣ + \frac{4}{9} ln ∣ x + 4 ∣ + \frac{31}{72} ln ∣ x - 5 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 2 x - 4}{( x + 3 ) ( x + 4 ) ( x - 5 )} : \frac{1}{8 ( x + 3 )} + \frac{4}{9 ( x + 4 )} + \frac{31}{72 ( x - 5 )}

= \int \frac{1}{8 ( x + 3 )} + \frac{4}{9 ( x + 4 )} + \frac{31}{72 ( x - 5 )} 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{4}{9 ( x + 4 )} d x + \int \frac{31}{72 ( x - 5 )} d x

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

\int \frac{4}{9 ( x + 4 )} d x = \frac{4}{9} ln ∣ x + 4 ∣

\int \frac{31}{72 ( x - 5 )} d x = \frac{31}{72} ln ∣ x - 5 ∣

= \frac{1}{8} ln ∣ x + 3 ∣ + \frac{4}{9} ln ∣ x + 4 ∣ + \frac{31}{72} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x + 3 ∣ + \frac{4}{9} ln ∣ x + 4 ∣ + \frac{31}{72} ln ∣ x - 5 ∣ + C

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What is the integral of (x^2+2x-4)/((x+3)(x+4)(x-5)) ?
The integral of (x^2+2x-4)/((x+3)(x+4)(x-5)) is 1/8 ln|x+3|+4/9 ln|x+4|+31/72 ln|x-5|+C