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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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\int \frac{4}{x ^{3} - 8} d x

\int (5 x^{3} - 3 x^{2} + 2) d x

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