What is the integral of (x^2+3x-12)/(x^3+3x^2-28x) ?

The integral of (x^2+3x-12)/(x^3+3x^2-28x) is 3/7 ln|x|+4/11 ln|x-4|+16/77 ln|x+7|+C

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integral of (x^2+3x-12)/(x^3+3x^2-28x)

\int \frac{x ^{2} + 3 x - 12}{x ^{3} + 3 x ^{2} - 28 x} d x

\frac{3}{7} ln ∣ x ∣ + \frac{4}{11} ln ∣ x - 4 ∣ + \frac{16}{77} ln ∣ x + 7 ∣ + C

Solution steps

\int \frac{x ^{2} + 3 x - 12}{x ^{3} + 3 x ^{2} - 28 x} d x

Take the partial fraction of

\frac{x ^{2} + 3 x - 12}{x ^{3} + 3 x ^{2} - 28 x} : \frac{3}{7 x} + \frac{4}{11 ( x - 4 )} + \frac{16}{77 ( x + 7 )}

= \int \frac{3}{7 x} + \frac{4}{11 ( x - 4 )} + \frac{16}{77 ( x + 7 )} 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{3}{7 x} d x + \int \frac{4}{11 ( x - 4 )} d x + \int \frac{16}{77 ( x + 7 )} d x

\int \frac{3}{7 x} d x = \frac{3}{7} ln ∣ x ∣

\int \frac{4}{11 ( x - 4 )} d x = \frac{4}{11} ln ∣ x - 4 ∣

\int \frac{16}{77 ( x + 7 )} d x = \frac{16}{77} ln ∣ x + 7 ∣

= \frac{3}{7} ln ∣ x ∣ + \frac{4}{11} ln ∣ x - 4 ∣ + \frac{16}{77} ln ∣ x + 7 ∣

Add a constant to the solution

= \frac{3}{7} ln ∣ x ∣ + \frac{4}{11} ln ∣ x - 4 ∣ + \frac{16}{77} ln ∣ x + 7 ∣ + C

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What is the integral of (x^2+3x-12)/(x^3+3x^2-28x) ?
The integral of (x^2+3x-12)/(x^3+3x^2-28x) is 3/7 ln|x|+4/11 ln|x-4|+16/77 ln|x+7|+C