[tex]\int\limits_{0}^{2} \frac{x}{(x {}^{2} + 1) {}^{2} } dx \\ \\ u = x {}^{2} + 1 \: \to \: \frac{du}{dx} = 2x \: \to \: \frac{du}{2} = xdx \\ \\ \int\limits_{0}^{2} \frac{ \frac{du}{2} }{u {}^{2} } \: \to \: \frac{1}{2} \int\limits_{0}^{2} \frac{du}{u {}^{2} } \: \to \: \frac{1}{2} \left( - \frac{1}{u} \right)_{0}^{2} \\ \\ \frac{1}{2} . \left[ - \frac{1}{x {}^{2} + 1 } \right] \: \to \: \frac{1}{2} . \left[ - \frac{1}{2 {}^{2} + 1 } + \frac{1}{0 {}^{2} + 1 } \right] \\ \\ \frac{1}{2} . \left[ - \frac{1}{5} + 1 \right] \: \to \: \frac{1}{2} . \frac{4}{5} \: \to \: \boxed{ \boxed{ \boxed{\frac{2}{5} }}}[/tex]

[tex] \int \limits_{0}^{ \frac{\pi}{2} } x \sin(2x) \: dx \\ \\ u = x \: \to \: \: \frac{du}{dx} = 1 \: \to \: du = dx \\ \\ dv = \sin(2x) \: \to \: \int dv = \int \sin(2x) \: dx \\ \\ v = \int \sin(2x) \: dx \: \to \: \: h = 2x \: \to \: \frac{dh}{dx} = 2 \: \to \: \frac{dh}{2} = dx \\ \\ v = \int \sin(h) \: . \: \frac{dh}{2} \: \to \: v = \frac{1}{2} \int \sin(h) \: dh \: \to \: v = - \frac{ \cos(2x)}{2} [/tex]

[tex] \int u.dv = u.v - \int v.du \\ \\ \int x. \sin(2x) \: dx = x. \left( - \frac{ \cos(2x)}{2} \right) + \int \frac{ \cos(2x)}{2} dx \\ \int x. \sin(2x) \: dx = x. \left( - \frac{ \cos(2x)}{2} \right) + \frac{1}{2} \int \cos(2x) \: dx \\ \int x. \sin(2x) \: dx = x. \left( - \frac{ \cos(2x)}{2} \right) + \frac{ \sin(2x)}{4} \\ \\ \left[x. \left( - \frac{ \cos(2x)}{2} \right) + \frac{ \sin(2x)}{4} \right] _ {0}^{ \frac{\pi}{2} } \\ \\ \frac{\pi}{2} . \left( - \frac{ \cos(\pi)}{2} \right) + \frac{ \sin(\pi)}{4} \: \to \: \frac{\pi}{2} . \frac{1}{2} \: \to \:\boxed{\boxed{\boxed{\frac{\pi}{4} }}}[/tex]

Portanto a resposta final é:

[tex] \boxed{ \boxed{ \boxed{{\frac{\pi}{4} + \frac{2}{5}} }}}[/tex]