What is the integral of (x+3)/(x(x-1)(x+4)) ?

The integral of (x+3)/(x(x-1)(x+4)) is 4/5 ln|x-1|-1/20 ln|x+4|-3/4 ln|x|+C

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integral of (x+3)/(x(x-1)(x+4))

\int \frac{x + 3}{x ( x - 1 ) ( x + 4 )} d x

\frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣ + C

Solution steps

\int \frac{x + 3}{x ( x - 1 ) ( x + 4 )} d x

Expand

\frac{x + 3}{x ( x - 1 ) ( x + 4 )} : \frac{x}{x ( x - 1 ) ( x + 4 )} + \frac{3}{x ( x - 1 ) ( x + 4 )}

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{x}{x ( x - 1 ) ( x + 4 )} d x + \int \frac{3}{x ( x - 1 ) ( x + 4 )} d x

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

\int \frac{3}{x ( x - 1 ) ( x + 4 )} d x = 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣)

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣)

Simplify

\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣) : \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣

= \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣

Add a constant to the solution

= \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣ + C

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\int_{4}^{5} \frac{x}{x - 4} d x

What is the integral of (x+3)/(x(x-1)(x+4)) ?
The integral of (x+3)/(x(x-1)(x+4)) is 4/5 ln|x-1|-1/20 ln|x+4|-3/4 ln|x|+C