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

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

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

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

x - 4 ln ∣ x - 1 ∣ + ln ∣ x ∣ - 1 + C

Solution steps

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

Expand

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

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

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

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

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

= x - 1 + ln ∣ x - 1 ∣ - 4 ln ∣ x - 1 ∣ - (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

Simplify

= x - 4 ln ∣ x - 1 ∣ + ln ∣ x ∣ - 1

Add a constant to the solution

= x - 4 ln ∣ x - 1 ∣ + ln ∣ x ∣ - 1 + C

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