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

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

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

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

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

Solution steps

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

Complete the square

u^{2} - 2 u : (u - 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{v ^{2} - 1} d v

= \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

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

Substitute back

v = u - 1

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

Simplify

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

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

Add a constant to the solution

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

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