Confusing answer for $\int\frac{x^4+1}{x^4-1}dx$

Question

While prepairing for upcoming test I met this integral and was confused with the answer they got in the answer book for it. Please show me where I am wrong. $$\int\frac{x^4+1}{x^4-1}dx=\int\frac{x^4}{x^4-1}dx + \int\frac{1}{x^4-1}dx=\int\frac{x^4-1+1}{x^4-1}dx + \int\frac{1}{x^4-1}dx=\int dx + 2\times\int\frac{1}{x^4-1}dx=\int dx+\int\frac{1}{(x^2)^2-1}d(x^2)=x + \frac{1}{2}ln{ \left|\frac{x-1}{x+1}\right|} + C$$ But the book has the following answer: $$x + \frac{1}{2}ln{ \left|\frac{x-1}{x+1} \right|} - \arctan x + C $$ So my question is: where did they took $\arctan x$? I understand that the first move they could do is to transform denominator into $(x^2-1)(x^2+1)$ but I dont understand why we have different answers.

Answer 1

well, you can't just replace $dx$ with $d(x^2)$.

the calculation is $$\int\frac{1}{x^4-1}dx=\int\frac{1}{(x-1)(x+1)(x^2+1)}dx$$ and from here you continue with partial fractions decomposition.

if you still want the conversion of $dx$, then if $u=x^2$ then $dx=du/2x=du/2\sqrt u$, and so $$\int \frac{1}{x^4-1}dx=\int \frac{1}{2\sqrt{u}(u^2-1)}du$$ which is a problem in itself