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

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

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integral of 1/(y+y^2)

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

ln ∣ y ∣ - ln ∣ y + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{y + y ^{2}} : \frac{1}{y} - \frac{1}{y + 1}

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

Apply the Sum Rule:

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{1}{y + 1} d y = ln ∣ y + 1 ∣

= ln ∣ y ∣ - ln ∣ y + 1 ∣

Add a constant to the solution

= ln ∣ y ∣ - ln ∣ y + 1 ∣ + C

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