TITLE: Integral of $\frac{t}{t^4+2} dt$
QUESTION [5 upvotes]: The answer in the back of the calculus book is $$\frac{1}{2\sqrt2}\arctan \left( \frac{t^2}{\sqrt2} \right) + C$$
and I have no idea how they reached this answer.  My first guess was to try partial fractions but I don't think I can in this case.  I then tried u substitution using $u=t^2$ and $du=2t$, giving me 
$$\frac{1}{2}\int \frac{du}{u^2+2}$$
I thought that I'd be able integrate this to reach an answer like $\frac{1}{2} \ln |t^2 + 1| + C$ but that's of course not the case and I'm not sure why.  How should I approach this?

REPLY [2 votes]: $$
\int \frac{t}{t^4+2}\,dt = \frac 1 {2\sqrt{2}} \int \frac{1}{\left(\underbrace{{}\quad\dfrac{t^2}{\sqrt{2}}\quad{}}_{\large u}\right)^2+1} \left(\underbrace{\sqrt{2}\,t\,dt}_{\large du}\right)
$$