\begin{aligned}\int_{0}^{+\infty}{\frac{\arctan{x}}{\sqrt[4]{x}\left(1+x\right)}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{x}{\sqrt[4]{x}\left(1+x\right)\left(1+x^{2}y^{2}\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\int_{0}^{+\infty}{\frac{x}{\sqrt[4]{x}\left(1+x\right)\left(1+x^{2}y^{2}\right)}\,\mathrm{d}x}\,\mathrm{d}y}\end{aligned}
Making the change of variable $ \left\lbrace\begin{aligned}x&=u^{4}\\ \mathrm{d}x&=4u^{3}\,\mathrm{d}u\end{aligned}\right. $, we get the following : \begin{aligned}\int_{0}^{+\infty}{\frac{\arctan{x}}{\sqrt[4]{x}\left(1+x\right)}\,\mathrm{d}x}&=4\int_{0}^{1}{\int_{0}^{+\infty}{\frac{u^{6}}{\left(1+u^{4}\right)\left(1+u^{8}y^{2}\right)}\,\mathrm{d}u}\,\mathrm{d}y}\\ &=4\int_{0}^{1}{\int_{0}^{+\infty}{\left(\frac{u^{6}y^{2}+u^{2}}{\left(1+y^{2}\right)\left(1+u^{8}y^{2}\right)}-\frac{u^{2}}{\left(1+y^{2}\right)\left(1+u^{4}\right)}\right)\mathrm{d}u}\,\mathrm{d}y}\\ &=4\int_{0}^{1}{\int_{0}^{+\infty}{\frac{u^{6}y^{2}+u^{2}}{\left(1+y^{2}\right)\left(1+u^{8}y^{2}\right)}\,\mathrm{d}u}\,\mathrm{d}y}-4\int_{0}^{1}{\int_{0}^{+\infty}{\frac{u^{2}}{\left(1+y^{2}\right)\left(1+u^{4}\right)}\,\mathrm{d}u}\,\mathrm{d}y}\end{aligned}
By fixing $ y $, and applying the change of variable $ \left\lbrace\begin{aligned}u&=\frac{1}{\sqrt[4]{y}t}\\ \mathrm{d}u&=-\frac{\mathrm{d}t}{\sqrt[4]{y}t^{2}}\end{aligned}\right. $ Inside of the first integral, we get the following : \begin{aligned}\int_{0}^{+\infty}{\frac{\arctan{x}}{\sqrt[4]{x}\left(1+x\right)}\,\mathrm{d}x}&=4\int_{0}^{1}{\int_{0}^{+\infty}{\frac{y+t^{4}}{\sqrt[4]{y^{3}}\left(1+y^{2}\right)\left(1+t^{8}\right)}\,\mathrm{d}t}\,\mathrm{d}y}-4\int_{0}^{1}{\int_{0}^{+\infty}{\frac{u^{2}}{\left(1+y^{2}\right)\left(1+u^{4}\right)}\,\mathrm{d}u}\,\mathrm{d}y}\\ &\scriptsize =4\left(\int_{0}^{1}{\frac{\sqrt[4]{y}}{1+y^{2}}\,\mathrm{d}y}\right)\left(\int_{0}^{+\infty}{\frac{\mathrm{d}t}{1+t^{8}}}\right)+4\left(\int_{0}^{1}{\frac{\sqrt[4]{y}}{1+y^{2}}\,\mathrm{d}y}\right)\left(\int_{0}^{+\infty}{\frac{t^{4}}{1+t^{8}}\,\mathrm{d}t}\right)-4\left(\int_{0}^{1}{\frac{\mathrm{d}y}{1+y^{2}}}\right)\left(\int_{0}^{+\infty}{\frac{u^{2}}{1+u^{4}}\,\mathrm{d}u}\right)\end{aligned}
Finally by applying the change of variable $ \left\lbrace\begin{aligned}y&=\varphi^{4}\\ \mathrm{d}y&=4\varphi^{3}\,\mathrm{d}\varphi\end{aligned}\right. $ in the first and the second term we get : $$\scriptsize \int_{0}^{+\infty}{\frac{\arctan{x}}{\sqrt[4]{x}\left(1+x\right)}\,\mathrm{d}x}=16\left(\int_{0}^{1}{\frac{\varphi^{4}}{1+\varphi^{8}}\,\mathrm{d}\varphi}\right)\left(\int_{0}^{+\infty}{\frac{\mathrm{d}t}{1+t^{8}}}\right)+16\left(\int_{0}^{1}{\frac{\varphi^{4}}{1+\varphi^{8}}\,\mathrm{d}\varphi}\right)\left(\int_{0}^{+\infty}{\frac{t^{4}}{1+t^{8}}\,\mathrm{d}t}\right)-4\left(\int_{0}^{1}{\frac{\mathrm{d}y}{1+y^{2}}}\right)\left(\int_{0}^{+\infty}{\frac{u^{2}}{1+u^{4}}\,\mathrm{d}u}\right) $$
I shall leave the rest for you. I suppose you know how to solve $ \int\limits_{0}^{+\infty}{\frac{x^{m}}{1+x^{n}}\,\mathrm{d}x} $ : $$ \int_{0}^{\infty}{\frac{x^{a-1}}{1+x^{b}}\,\mathrm{d}x} = \frac{\pi}{b \sin{\left(\frac{\pi a}{b}\right)}}, \qquad 0 < a <b $$
Best Answer
You can apply l'Hôpital to any form $?/\infty$, where the question mark means any function that's differentiable where needed. On the other hand, it's not difficult to show that $$ \lim_{x\to\infty}\int_0^x(\arctan t)^2\,dt=\infty $$ so you can apply the standard form.
The fundamental theorem of calculus then allows you to compute $$ \lim_{x\to\infty}\frac{(\arctan x)^2}{x/\sqrt{x^2-1}} $$ and this is not indeterminate.