The integral sign has a 2 at the top and a 0 at the bottom
2/x^2+1 dx
You have $\displaystyle \int_{0}^{2}\dfrac{2}{x^2+1}\;{dx}$. Let $\displaystyle x = \tan\theta$, then $\displaystyle \dfrac{dx}{d\theta} = \sec^2\theta \Rightarrow dx = \left(d\theta\right){\sec^2\theta}$. When $\displaystyle x = 2$, $\displaystyle \tan\theta = 2 \Rightarrow \theta = \arctan\left(2\right) $. When $\displaystyle x = 0$, $\displaystyle \tan{\theta} = 0 \Rightarrow \theta = 0$ . So we have $\displaystyle \int_{x=0}^{x=2}\dfrac{2}{x^2+1}\;{dx} = \int_{\theta=0}^{\theta=\arctan(2)}\dfrac{2}{\tan^ 2\theta+1}{\sec^2\theta}\;{d\theta} = \int_{0}^{\arctan(2)}{2}\;{d\theta}$ $\displaystyle = \bigg[2\theta\bigg]_{0}^{\arctan(2)} = \boxed{2\arctan(2)}.$