Answer
\[ = \frac{4}{5}\]
Work Step by Step
\[\begin{gathered}
\hfill \\
\int_0^2 {\frac{{2x}}{{\,{{\left( {{x^2} + 1} \right)}^2}}}} \,dx \hfill \\
\hfill \\
Let\,\,u = {x^2} + 1\,\,\,{\text{which implies that}}\,\,\,\,\,du = 2xdx \hfill \\
\hfill \\
{\text{Changing limits of integration}} \hfill \\
\hfill \\
x = 2\,\,\,implies\,\,\,\,u = 5 \hfill \\
x = 0\,\,implies\,\,\,\,u = 1 \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
\int_1^5 {\frac{{du}}{{{u^2}}}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \,\,\left[ { - \frac{1}{u}} \right]_1^5 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= - \frac{1}{5} + \frac{1}{1} = \frac{4}{5} \hfill \\
\end{gathered} \]
