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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = y - 1

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

Simplify

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

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

Add a constant to the solution

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

3 y^{^{'}} + y = 0, y (0) = 0

n \to \infty lim (\frac{1}{e})

\int \frac{5 ( x ^{2} + 4 )}{x ( x ^{2} + 5 )} d x

x \to 0 lim (+ (sin (x)))

f (x) = x + 8, at x = 1

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