Answer
\[ = \frac{1}{2}\]
Work Step by Step
\[\begin{gathered}
\int_1^3 {\frac{2}{{{x^2} + 2x + 1}}\,\,dx} \hfill \\
\hfill \\
factor\,ing\,\,the\,\,denominator \hfill \\
\hfill \\
= \int_1^3 {\frac{2}{{\,{{\left( {x + 1} \right)}^2}}}} \,\,dx \hfill \\
\hfill \\
set\,x + 1 = t\,\,\,\,\,\, \to \,\,\,\,\,dx = dt \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \int_2^4 {\frac{2}{{{t^2}}}\,dt} \, \hfill \\
\hfill \\
\,integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \,\,\left[ { - \frac{2}{t}} \right]_2^4 = - 2\,\,\left[ {\frac{1}{4} - \frac{1}{2}} \right] = \frac{1}{2} \hfill \\
\end{gathered} \]
