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?