Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?  

There are two (three) ways to go. One, assume 
$$x^4+1=(x^2+ax+1)(x^2-ax+1)$$
You'll get that
$${x^4} + 1 = {x^4} + \left( {2 - {a^2}} \right){x^2} + 1$$
Then $a=\sqrt 2$ (or the other, by symmetry)
$${x^4} + 1 = {x^4} + 1 = \left( {{x^2} + \sqrt 2 x + 1} \right)\left( {{x^2} - \sqrt 2 x + 1} \right)$$
The other ${x^2} = \tan \theta $, but it might get messy, unless you know how to use the Weierstrass substitution for example. 
$$\int {\frac{{dx}}{{{x^4} + 1}}}  = \int {\frac{{\left( {{{\tan }^2}\theta  + 1} \right)d\theta }}{{{{\tan }^2}\theta  + 1}}} \frac{1}{{2\sqrt {\tan \theta } }} = \int {\sqrt {\frac{{\cos\theta }}{{\sin\theta }}} \frac{{d\theta }}{2}} $$
$$\int {\sqrt {\frac{{\frac{{1 - {u^2}}}{{1 + {u^2}}}}}{{\frac{{2u}}{{1 + {u^2}}}}}} \frac{{du}}{{1 + {u^2}}}}  = \int {\sqrt {\frac{{1 - {u^2}}}{{2u}}} \frac{{du}}{{1 + {u^2}}}} $$
However, Chandrasekar's is the best way to go, if you can figure it out.
