Find $\int \frac{4x}{x^4-1}dx$

Question

Find $\int \frac{4x}{x^4-1} dx$ using partial fractions: $\frac{4x}{x^4-1}=\frac{4x}{(x^2+1)(x^2-1)}=\frac{4x}{(x^2+1)(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{Cx+D}{x^2+1}$ we see that therefore $A=1, B=1, C=-2, D=0$ which gives us the following indefinite integral: $\int \frac{1}{x+1}+\frac{1}{x-1}+\frac{-2x}{x^2+1} dx$ The first two terms I come up with $\ln |x+1|+\ln |x-1|$ And for the final term I u-substitute: $u=x^2, -du=-2xdx$ therefore the last integral becomes $-\int \frac{du}{u+1}$ which is $-\ln |u+1|$ All told I have $$\int \frac{4x}{x^4-1} dx=\ln |x+1|+\ln |x-1|-\ln |x^2+1| +c$$ No? (Apologies now for errors in typesetting)

Answer 1

Assume $x^2=p$ and you get $x^4-1=p^2-1$ and $2x\cdot dx=dp$. Thus your integral is equivalent to $2\int \frac{1}{p^2-1} \cdot dp$