$\begin{align}  & \text{Integrating without the limits} \\ & \int{(2x{{\sec }^{2}}{{x}^{2}}+\tan x)dx}=\int{2x{{\sec }^{2}}{{x}^{2}}dx}+\int{\tan xdx} \\ & \int{2x{{\sec }^{2}}{{x}^{2}}dx}=2x\tan {{x}^{2}}-\int{\tan {{x}^{2}}(2dx)} \\ & \int{2x{{\sec }^{2}}{{x}^{2}}dx}=2x\tan {{x}^{2}}-2\int{\tan {{x}^{2}}dx} \\ & \int{2x{{\sec }^{2}}{{x}^{2}}dx=2x\tan {{x}^{2}}}+2\ln \cos {{x}^{2}}----(1) \\ & \int{\tan xdx}=-\ln \cos x \\ & \text{Adding }(1)\text{ and }(2)\text{ together} \\ & \int{(2x{{\sec }^{2}}{{x}^{2}}+\tan x)dx}=2x\tan {{x}^{2}}+2\ln \cos {{x}^{2}}-\ln \cos x \\ & \int_{0}^{\tfrac{\pi }{4}}{(2x{{\sec }^{2}}{{x}^{2}}+\tan x)dx}=[2x\tan {{x}^{2}}+2\ln \cos {{x}^{2}}-\ln \cos x]_{0}^{\tfrac{\pi }{2}} \\ & =\left( 2(\tfrac{\pi }{4})\tan {{(\tfrac{\pi }{4})}^{2}}+2\ln \cos {{(\tfrac{\pi }{4})}^{2}}-\ln \cos (\tfrac{\pi }{4}) \right)-(0+0-0) \\ & =\frac{\pi }{2}+4\ln \tfrac{1}{\sqrt{2}}-\ln \tfrac{1}{\sqrt{2}}=\frac{\pi }{2}+3\ln \frac{1}{\sqrt{2}} \\\end{align}$