What is #int x^4/(x^4-1)dx#?

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Answer 1 · 2015-11-06T13:58:12

#[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C#

#intx^4/(x^4-1)dx = int(x^4-1+1)/(x^4-1)dx = int1dx + int1/(x^4-1)dx#

Looking at #int1/(x^4-1)dx#

you have #x^4-1=(x^2+1)(x^2-1) = (x^2+1)(x-1)(x+1)#

So

#int1/((x^2+1)(x+1)(x-1))dx#

You need to do partial fraction, you get

#-1/2int1/((x^2+1))dx-1/4int1/((x+1))dx+1/4int1/((x-1))dx#

#[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C#