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

The integral of 1/(y^3-y) is -ln|y|+1/2 ln|y+1|+1/2 ln|y-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 + 1 ∣ + \frac{1}{2} ln ∣ y - 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{1}{2 ( y + 1 )} + \frac{1}{2 ( y - 1 )}

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

\int \frac{1}{y} d y = ln ∣ y ∣

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

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

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

Add a constant to the solution

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

h (x) = \frac{7}{x}, at x = 5

\frac{d}{d x} (- \frac{1}{2} e^{- \frac{x}{4}})

x \to 0 + lim (x lo g_{e} (x))

- 8 (64 e^{- 4 x^{2}} x^{3} - 24 e^{- 4 x^{2}} x)

\frac{t ^{2}}{2}

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