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

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

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integral of (2y)/(-2xy+x^2)

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \frac{1}{x} \cdot \int \frac{u - x}{4 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{x} \cdot \frac{1}{4} \cdot \int \frac{u - x}{u} d u

Expand

\frac{u - x}{u} : 1 - \frac{x}{u}

= 2 \cdot \frac{1}{x} \cdot \frac{1}{4} \cdot \int 1 - \frac{x}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{x} \cdot \frac{1}{4} (\int 1 d u - \int \frac{x}{u} d u)

\int 1 d u = u

\int \frac{x}{u} d u = x ln ∣ u ∣

= 2 \cdot \frac{1}{x} \cdot \frac{1}{4} (u - x ln ∣ u ∣)

Substitute back

u = x - 2 y

= 2 \cdot \frac{1}{x} \cdot \frac{1}{4} (x - 2 y - x ln ∣ x - 2 y ∣)

Simplify

2 \cdot \frac{1}{x} \cdot \frac{1}{4} (x - 2 y - x ln ∣ x - 2 y ∣) : \frac{1}{2} - \frac{y}{x} - \frac{1}{2} ln ∣ x - 2 y ∣

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

Add a constant to the solution

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

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