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

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

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

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

5 ln ∣ x ∣ + \frac{4}{x} - 5 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{5}{x} - \frac{4}{x ^{2}} - \frac{5}{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{5}{x} d x - \int \frac{4}{x ^{2}} d x - \int \frac{5}{x + 1} d x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

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

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

= 5 ln ∣ x ∣ - (- \frac{4}{x}) - 5 ln ∣ x + 1 ∣

Simplify

= 5 ln ∣ x ∣ + \frac{4}{x} - 5 ln ∣ x + 1 ∣

Add a constant to the solution

= 5 ln ∣ x ∣ + \frac{4}{x} - 5 ln ∣ x + 1 ∣ + C

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