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

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

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integral of 1/(u^2-u)

\int \frac{1}{u ^{2} - u} d u

- ln ∣ u ∣ + ln ∣ u - 1 ∣ + C

Solution steps

\int \frac{1}{u ^{2} - u} d u

Take the partial fraction of

\frac{1}{u ^{2} - u} : - \frac{1}{u} + \frac{1}{u - 1}

= \int - \frac{1}{u} + \frac{1}{u - 1} d u

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{1}{u} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

Add a constant to the solution

= - ln ∣ u ∣ + ln ∣ u - 1 ∣ + C

x \to - 2 lim (\frac{x ^{3} - 8}{x - 2})

\frac{\partial}{\partial y} (x^{2} + 4 y)

(7, - 5), (- 2, - 5)

y = 5^{6 x^{2}}

n = 2 \sum \infty \frac{1}{( n + 1 ) ^{2}}

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