What is the integral of 1/(y^3+y) ?

The integral of 1/(y^3+y) is ln|y|-1/2 ln|y^2+1|+C

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integral of 1/(y^3+y)

\int \frac{1}{y ^{3} + y} d y

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

Solution steps

\int \frac{1}{y ^{3} + y} d y

Take the partial fraction of

\frac{1}{y ^{3} + y} : \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

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

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

\frac{x ^{3} - 1}{x ^{2}}

\int \frac{3 x - 1}{x ^{2} + 9} d x

\frac{d}{d x} (ln (g (x)))

What is the integral of 1/(y^3+y) ?
The integral of 1/(y^3+y) is ln|y|-1/2 ln|y^2+1|+C