$\int \frac{4 x}{1 - x^{4}} d x$ $int (4x)/(1-x^4)dx$ =?

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$\int \frac{4 x}{1 - x^{4}} d x$ $int (4x)/(1-x^4)dx$ =?

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Answer 1 · 2016-06-01T22:14:33

Notice that

$4 (\frac{x}{1 - x^{4}}) = 2 \frac{x}{(x^{2} + 1)} - \frac{1}{(x - 1)} - \frac{1}{(x + 1)}$ $4(x/(1-x^4))=2x/( (x^2+1))-1/( (x-1))-1/( (x+1))$

Hence

$\int \frac{x}{1 - x^{4}} d x = 2 \int \frac{x}{(x^{2} + 1)} d x - \int \frac{1}{(x - 1)} d x - \int \frac{1}{(x + 1)} d x = log (x^{2} + 1) - log (x - 1) - log (x + 1) + c = log (x^{2} + 1) - log (x^{2} - 1) + c$ $int x/(1-x^4)dx=2int x/( (x^2+1))dx-int1/( (x-1))dx-int1/( (x+1))dx=log(x^2+1)-log(x-1)-log(x+1)+c= log(x^2+1)-log(x^2-1)+c$

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$\int \frac{4 x}{1 - x^{4}} d x$ $int (4x)/(1-x^4)dx$ =?

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