Trying to find a function whose derivative is $\cfrac{4x^3}{x^4+1}$ any ideas? The closet I can get is
$\arctan(x^4) $
the derivative of $\arctan(x^4)$ is $\dfrac{4x^3}{x^8 + 1} ...$
Any tips?

Hint: Note that the derivative of $x^4 + 1\;$ is $4x^3$.
If we let $f(x) = x^4 + 1$, then we know $f'(x) = 4x^3$. 
Then note that $$\frac{4x^3}{x^4 + 1} = \dfrac{f'(x)}{f(x)}\tag{1}$$
And all integrals of the form $\displaystyle \int \dfrac{f'x}{f(x)}\,dx $ evaluate as  $$\int \dfrac{f'x}{f(x)}\,dx = \ln|f(x)| + C \tag{where C is some constant}$$


*

*Just note that $\;\frac{d}{dx}(\ln|f(x)| + C) = \dfrac{f'(x)}{f(x)}$ (by the chain rule).


If you already know how to integrate, you'll find we can integrate by substitution: 
Let $u = f(x) =x^4 + 1$, then $f'(x) = du/dx = 4x^3 \implies \; du = 4x^3 \,dx$.
$$
\begin{align} \int \dfrac{4x^3\,dx}{x^4 + 1}\tag{By $(1)$}
& = \int \dfrac{du}{u} = \ln|u| + C \\ \\
& = \ln(x^4 + 1) + C \\ \\
\end{align}
$$
(because $x^4 + 1 \gt 0$, we don't need the absolute value sign: $|x^4 + 1|)$
If you haven't learned integration, note that $\; \dfrac{4x^3}{x^4 + 1}\;$ is precisely the derivative of $\ln(x^4 + 1)$. 
You can check for yourself: using the chain rule, we get $$\frac d{dx}(\ln(x^4 + 1)) = \dfrac d{dx}(x^4 + 1)\cdot \dfrac {1}{x^4 + 1} = 4x^3 \cdot \dfrac{1}{x^4 + 1}$$
