What is the integral of ((2x^2-1))/(x-x^3) ?

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

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integral of ((2x^2-1))/(x-x^3)

\int \frac{( 2 x ^{2} - 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{2 x ^{2} - 1}{x - x ^{3}} d x

Take the partial fraction of

\frac{2 x ^{2} - 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

\frac{d}{d x} (sin (tan (x^{5})))

\frac{d y}{d x} = 3 - 0.0003 y

\frac{d}{d x} (3 x^{2} - 6 x + 5)

\int (\frac{1}{y} - 1) d y

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