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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 2 x

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

Add a constant to the solution

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

\frac{d}{d t} (2)

\int x + x e^{x} d x

a re a x^{3} + x^{2} - 6 x, 6 x

a re a f (x) = x^{2} - 4, [- 4, 3]

\int x - 5 d x

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