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

The integral of (x-4)/(x(x-1)^2) is 3/(x-1)-4ln|x|+4ln|x-1|+C

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{4}{x ( x - 1 ) ^{2}} d x = 4 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

Simplify

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

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

Add a constant to the solution

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

x \to 2 lim (\frac{x - 2}{x - 2})

\int (yz e^{x z}) d x

t \frac{d y}{d t} + 3 y = 9 t

\int (x + 3) x d x

What is the integral of (x-4)/(x(x-1)^2) ?
The integral of (x-4)/(x(x-1)^2) is 3/(x-1)-4ln|x|+4ln|x-1|+C