What is the integral of ((y+2))/(y^2+4y) ?

The integral of ((y+2))/(y^2+4y) is 1/2 ln|y^2+4y|+C

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integral of ((y+2))/(y^2+4y)

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

\frac{1}{2} ln y^{2} + 4 y + C

Solution steps

\int \frac{y + 2}{y ^{2} + 4 y} d y

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 = y^{2} + 4 y

= \frac{1}{2} ln y^{2} + 4 y

Add a constant to the solution

= \frac{1}{2} ln y^{2} + 4 y + C

What is the integral of ((y+2))/(y^2+4y) ?
The integral of ((y+2))/(y^2+4y) is 1/2 ln|y^2+4y|+C