Answer
\[ = - \frac{1}{{2\,\left( {{x^2} + 1} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{{x^4} + 2{x^2} + 1}}dx} \hfill \\
\hfill \\
factor\,\,the\,\,deno\min ator \hfill \\
\hfill \\
= \int_{}^{} {\frac{x}{{\,{{\left( {{x^2} + 1} \right)}^2}}}dx} \hfill \\
\hfill \\
set\,\,\,\left( {{x^2} + 1} \right) = t{\text{ then }}2xdx = dt\,\, \to \,\,xdx = \frac{{dt}}{2} \hfill \\
\hfill \\
integral\,becomes \hfill \\
\hfill \\
= \int_{}^{} {\frac{x}{{\,{{\left( {{x^2} + 1} \right)}^2}}}dx} = \int_{}^{} {\frac{{dt}}{{2{t^2}}}} \hfill \\
\hfill \\
{\text{Integrate using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }}} \hfill \\
\hfill \\
= - \frac{1}{{2t}} + C \hfill \\
\hfill \\
substituting\,back\,t = \,\left( {{x^2} + 1} \right) \hfill \\
\hfill \\
= - \frac{1}{{2\,\left( {{x^2} + 1} \right)}} + C \hfill \\
\end{gathered} \]
