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

The integral of (x-5)/(x^2(x+1)) is 6ln|x|+5/x-6ln|x+1|+C

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

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

6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x - 5}{x ^{2} ( x + 1 )} : \frac{6}{x} - \frac{5}{x ^{2}} - \frac{6}{x + 1}

= \int \frac{6}{x} - \frac{5}{x ^{2}} - \frac{6}{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{6}{x} d x - \int \frac{5}{x ^{2}} d x - \int \frac{6}{x + 1} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

= 6 ln ∣ x ∣ - (- \frac{5}{x}) - 6 ln ∣ x + 1 ∣

Simplify

= 6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣

Add a constant to the solution

= 6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣ + C

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What is the integral of (x-5)/(x^2(x+1)) ?
The integral of (x-5)/(x^2(x+1)) is 6ln|x|+5/x-6ln|x+1|+C