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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 1

Add a constant to the solution

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

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