What is the integral of (x-60)/(x^2(x+4)) ?

The integral of (x-60)/(x^2(x+4)) is 4ln|x|+(15)/x-4ln|x+4|+C

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integral of (x-60)/(x^2(x+4))

\int \frac{x - 60}{x ^{2} ( x + 4 )} d x

4 ln ∣ x ∣ + \frac{15}{x} - 4 ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{x - 60}{x ^{2} ( x + 4 )} d x

Take the partial fraction of

\frac{x - 60}{x ^{2} ( x + 4 )} : \frac{4}{x} - \frac{15}{x ^{2}} - \frac{4}{x + 4}

= \int \frac{4}{x} - \frac{15}{x ^{2}} - \frac{4}{x + 4} 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{4}{x} d x - \int \frac{15}{x ^{2}} d x - \int \frac{4}{x + 4} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{15}{x ^{2}} d x = - \frac{15}{x}

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

= 4 ln ∣ x ∣ - (- \frac{15}{x}) - 4 ln ∣ x + 4 ∣

Simplify

= 4 ln ∣ x ∣ + \frac{15}{x} - 4 ln ∣ x + 4 ∣

Add a constant to the solution

= 4 ln ∣ x ∣ + \frac{15}{x} - 4 ln ∣ x + 4 ∣ + C

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What is the integral of (x-60)/(x^2(x+4)) ?
The integral of (x-60)/(x^2(x+4)) is 4ln|x|+(15)/x-4ln|x+4|+C