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

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

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

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

- ln ∣ x ∣ + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

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

Substitute back

u = x - 1

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

Simplify

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

= - ln ∣ x ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ x - 2 ∣ + C

\frac{d}{d x} (\frac{x}{lo g _{10} ( x )})

y = \frac{1}{lo g _{4} ( x )}

\int_{0}^{4} \frac{2}{x ^{2}} d x

\int_{- 1}^{1} 2 x^{2} d x

\frac{d}{d x} (\frac{- 20}{( 2 + x ) ^{2}})

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