What is the integral of (x+2)/(1-x^4) ?

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

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integral of (x+2)/(1-x^4)

\int \frac{x + 2}{1 - x ^{4}} d x

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

Solution steps

\int \frac{x + 2}{1 - x ^{4}} d x

Expand

\frac{x + 2}{1 - x ^{4}} : \frac{x}{1 - x ^{4}} + \frac{2}{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}{1 - x ^{4}} d x + \int \frac{2}{1 - x ^{4}} d x

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

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

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

Add a constant to the solution

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

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