What is the integral of 1/(y^3-4y) ?

The integral of 1/(y^3-4y) is -1/4 ln|y|+1/8 ln|y+2|+1/8 ln|y-2|+C

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integral of 1/(y^3-4y)

\int \frac{1}{y ^{3} - 4 y} d y

- \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 2 ∣ + C

Solution steps

\int \frac{1}{y ^{3} - 4 y} d y

Factor

y^{3} - 4 y : - (- y^{3} + 4 y)

= \int \frac{1}{- ( - y ^{3} + 4 y )} d y

Take the constant out:

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

= - \int \frac{1}{- y ^{3} + 4 y} d y

Take the partial fraction of

\frac{1}{- y ^{3} + 4 y} : \frac{1}{4 y} - \frac{1}{8 ( y + 2 )} - \frac{1}{8 ( y - 2 )}

= - \int \frac{1}{4 y} - \frac{1}{8 ( y + 2 )} - \frac{1}{8 ( y - 2 )} 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}{4 y} d y - \int \frac{1}{8 ( y + 2 )} d y - \int \frac{1}{8 ( y - 2 )} d y)

\int \frac{1}{4 y} d y = \frac{1}{4} ln ∣ y ∣

\int \frac{1}{8 ( y + 2 )} d y = \frac{1}{8} ln ∣ y + 2 ∣

\int \frac{1}{8 ( y - 2 )} d y = \frac{1}{8} ln ∣ y - 2 ∣

= - (\frac{1}{4} ln ∣ y ∣ - \frac{1}{8} ln ∣ y + 2 ∣ - \frac{1}{8} ln ∣ y - 2 ∣)

Simplify

= - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 2 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 2 ∣ + C

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What is the integral of 1/(y^3-4y) ?
The integral of 1/(y^3-4y) is -1/4 ln|y|+1/8 ln|y+2|+1/8 ln|y-2|+C