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

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

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integral of 1/(y(y+2))

\int \frac{1}{y ( y + 2 )} d y

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

Solution steps

\int \frac{1}{y ( y + 2 )} d y

Take the partial fraction of

\frac{1}{y ( y + 2 )} : \frac{1}{2 y} - \frac{1}{2 ( y + 2 )}

= \int \frac{1}{2 y} - \frac{1}{2 ( y + 2 )} 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}{2 y} d y - \int \frac{1}{2 ( y + 2 )} d y

\int \frac{1}{2 y} d y = \frac{1}{2} ln ∣ y ∣

\int \frac{1}{2 ( y + 2 )} d y = \frac{1}{2} ln ∣ y + 2 ∣

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

Add a constant to the solution

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

- 2 sin (x)

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

\int \frac{8 x}{2 x ^{2} + 1} d x

y = (x + 1)^{3}

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

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