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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.

dimaastronom
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    Your evaluation of $ \int \frac{\mathrm{d}x}{x^4-1}$ is wrong since $\mathrm{d}(x^2) = 2x \mathrm{d}x$, an $x$ you do not have in the integrand. To check your answer try differentiating $\frac{1}{2} \log \left| \frac{x-1}{x+1}\right|$ and see what you get. – N3buchadnezzar Mar 08 '15 at 20:54
  • yeah, you are right. Just noticed that – dimaastronom Mar 08 '15 at 20:55

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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

tzoorp
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