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

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

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integral of 1/(y-y^3)

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

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

Solution steps

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

Take the partial fraction of

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

8 cos (t)

y^{^{'}} + 2 y = 3 e^{2 x}

\int_{0}^{4} 2 π (x) (4 x - x^{2}) d x

\frac{d}{d x} (\frac{1}{x ^{- 5}})

2 \times 2 + 6 \times 1/22 \times 2

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