Answer
$$\frac{1}{\sqrt 2} \tan^{-1}(4\sqrt 2).$$
Work Step by Step
We have
$$\int_0^4\frac{dx}{2x^2+1}=\frac{1}{\sqrt 2}\int_0^4\frac{d(\sqrt 2x)}{(\sqrt{2}x)^2+1}\\
=\frac{1}{\sqrt 2} \tan^{-1}(\sqrt 2x)|_0^4=\frac{1}{\sqrt 2} \tan^{-1}(4\sqrt 2).$$
