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

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

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Answer 1 · 2015-11-06T13:58:12

$[x] - \frac{1}{2} [arctan (x)] - \frac{1}{4} [ln | x + 1 |] + \frac{1}{4} [ln | x - 1 |] + C$ $[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C$

Explanation:

$\int \frac{x^{4}}{x^{4} - 1} d x = \int \frac{x^{4} - 1 + 1}{x^{4} - 1} d x = \int 1 d x + \int \frac{1}{x^{4} - 1} d x$ $intx^4/(x^4-1)dx = int(x^4-1+1)/(x^4-1)dx = int1dx + int1/(x^4-1)dx$

Looking at $\int \frac{1}{x^{4} - 1} d x$ $int1/(x^4-1)dx$

you have $x^{4} - 1 = (x^{2} + 1) (x^{2} - 1) = (x^{2} + 1) (x - 1) (x + 1)$ $x^4-1=(x^2+1)(x^2-1) = (x^2+1)(x-1)(x+1)$

So

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

You need to do partial fraction, you get

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

$[x] - \frac{1}{2} [arctan (x)] - \frac{1}{4} [ln | x + 1 |] + \frac{1}{4} [ln | x - 1 |] + C$ $[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C$

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

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