What is the integral from 0 to x of (x^4)/(1-x^4) ?

The integral from 0 to x of (x^4)/(1-x^4) is -x+1/2 arctan(x)+1/4 ln|x+1|-1/4 ln|x-1|

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integral from 0 to x of (x^4)/(1-x^4)

\int_{0}^{x} \frac{x ^{4}}{1 - x ^{4}} d x

- x + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

Solution steps

\int_{0}^{x} \frac{x ^{4}}{1 - x ^{4}} d x

Take the partial fraction of

\frac{x ^{4}}{1 - x ^{4}} : - 1 + \frac{1}{2 ( x ^{2} + 1 )} + \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x - 1 )}

= \int_{0}^{x} - 1 + \frac{1}{2 ( x ^{2} + 1 )} + \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( 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_{0}^{x} 1 d x + \int_{0}^{x} \frac{1}{2 ( x ^{2} + 1 )} d x + \int_{0}^{x} \frac{1}{4 ( x + 1 )} d x - \int_{0}^{x} \frac{1}{4 ( x - 1 )} d x

\int_{0}^{x} 1 d x = x

\int_{0}^{x} \frac{1}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} arctan (x)

\int_{0}^{x} \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int_{0}^{x} \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

= - x + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

\int_{0}^{e} \frac{x}{x ^{2} + e} d x

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What is the integral from 0 to x of (x^4)/(1-x^4) ?
The integral from 0 to x of (x^4)/(1-x^4) is -x+1/2 arctan(x)+1/4 ln|x+1|-1/4 ln|x-1|