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

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

Popular >

integral of 1/(y^2+y)

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

ln ∣ y ∣ - ln ∣ y + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

\frac{x}{x ^{2} + 49}

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{2} + 9})

\int 3 e^{6} d x

\frac{\partial}{\partial x} (x^{3} y z^{2} + 3 yz)

f (x) = (x^{3} - 7) (3 x^{2} + 5)

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