$\huge \int_{0} ^{4} \frac{\d x}{x^2+16} $ $\huge \begin{align} \int \frac{\d x}{ x^2+16 } &= \frac{1}{4} \arctan{\frac{x}{4}} + C\\ \int_{0} ^{4} \frac{\d x}{x^2+16} &= \frac{1}{4} \pa { \arctan{1} - \arctan 0} \\ &= \frac{\arctan(1)}{4} \end{align} $