What is the integral of y/(x(x+y)) ?

The integral of y/(x(x+y)) is -ln((|y+x|}{|y|})+ln|\frac{y+x)/y-1|+C

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integral of y/(x(x+y))

\int \frac{y}{x ( x + y )} d x

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

Solution steps

\int \frac{y}{x ( x + y )} d x

Take the constant out:

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

= y \cdot \int \frac{1}{( y + x ) x} d x

Apply u-substitution

: \int \frac{1}{( u - y ) u} d u

= y \cdot \int \frac{1}{( u - y ) u} d u

u - y = y (\frac{u}{y} - 1)

= y \cdot \int \frac{1}{y ( \frac{u}{y} - 1 ) u} d u

Take the constant out:

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

= y \frac{1}{y} \cdot \int \frac{1}{( \frac{u}{y} - 1 ) u} d u

Apply u-substitution

: \int \frac{1}{v ( v - 1 )} d v

= y \frac{1}{y} \cdot \int \frac{1}{v ( v - 1 )} d v

Take the partial fraction of

\frac{1}{v ( v - 1 )} : - \frac{1}{v} + \frac{1}{v - 1}

= y \frac{1}{y} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

Apply the Sum Rule:

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

= y \frac{1}{y} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x y^{^{'}} + y = e^{x}, y (1) = 6

\int \frac{9}{6 + 9 x} d x

\int (2^{x} + e^{π}) d x

e^{7}

d x + e^{3} x d y = 0

What is the integral of y/(x(x+y)) ?
The integral of y/(x(x+y)) is -ln((|y+x|}{|y|})+ln|\frac{y+x)/y-1|+C