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

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

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

integral

\frac{( x ^{3} - 3 x )}{( 1 - x ^{4} )}

- \frac{1}{4} ln 1 - x^{4} - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

Solution steps

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

Expand

\frac{x ^{3} - 3 x}{1 - x ^{4}} : \frac{x ^{3}}{1 - x ^{4}} - \frac{3 x}{1 - x ^{4}}

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

\int \frac{x ^{3}}{1 - x ^{4}} d x = - \frac{1}{4} ln 1 - x^{4}

\int \frac{3 x}{1 - x ^{4}} d x = \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1)

= - \frac{1}{4} ln 1 - x^{4} - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1)

Add a constant to the solution

= - \frac{1}{4} ln 1 - x^{4} - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

\frac{4}{( x ^{2} - 4 )}

\frac{x ^{3}}{5}

cos^{2} (x) - 1

\frac{x}{x + 3}

e^{\frac{1}{2}} x^{2}

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