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

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

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integral of 1/(x-x^3)

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{x - x ^{3}} : \frac{1}{x} - \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

x = y^{2} - 15, x = 2 y

t \to 0 lim (18 (1.15)^{t})

f (x) = x^{4} + 3 e^{x}, (0, 3)

\frac{\partial}{\partial y} ((x y)^{\frac{1}{2}})

\frac{1}{8 y ^{2}}

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