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

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

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

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

2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

ln ∣ x - 1 ∣ + 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{x}

= 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{x} + C

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