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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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