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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

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

= ln ∣ y ∣ - \frac{1}{2} ln y^{2} + 1

Add a constant to the solution

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

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