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

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

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

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

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

Solution steps

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

Expand

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{17}{4}

\frac{d}{d x} (x^{\frac{3}{2}} + 4)

y = sin (cot (3 π x))

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 0

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