What is the integral of 1/(y*(2+y^3)) ?

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

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integral of 1/(y*(2+y^3))

\int \frac{1}{y \cdot ( 2 + y ^{3} )} d y

\frac{1}{3} (\frac{1}{2} ln y^{3} - \frac{1}{2} ln y^{3} + 2) + C

Solution steps

\int \frac{1}{y ( 2 + y ^{3} )} d y

Apply u-substitution

: \int \frac{1}{3 u ( 2 + u )} d u

= \int \frac{1}{3 u ( 2 + u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{u ( 2 + u )} d u

Take the partial fraction of

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

= \frac{1}{3} \cdot \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

= \frac{1}{3} (\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}{3} (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = y^{3}

= \frac{1}{3} (\frac{1}{2} ln y^{3} - \frac{1}{2} ln y^{3} + 2)

Add a constant to the solution

= \frac{1}{3} (\frac{1}{2} ln y^{3} - \frac{1}{2} ln y^{3} + 2) + C

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