Answer
\[\frac{1}{4}\ln \,\left( {{x^4} + 4{x^2} + 7} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^3} + 2x}}{{{x^4} + 4{x^2} + 7}}dx} \hfill \\
Let\,\,u = {x^4} + 4{x^2} + 7 \hfill \\
So\,\,that \hfill \\
du = \,\left( {4{x^3} + 8x} \right)dx \hfill \\
du = 4\,\left( {{x^3} + 2x} \right)dx \hfill \\
\int_{}^{} {\frac{{{x^3} + 2x}}{{{x^4} + 4{x^2} + 7}}dx = \frac{1}{4}\int_{}^{} {\frac{{4\,\left( {{x^3} + 2x} \right)dx}}{{{x^4} + 4{x^2} + 7}}} } \hfill \\
Integrating \hfill \\
\frac{1}{4}\int_{}^{} {\frac{{du}}{u} = \frac{1}{4}\ln \left| u \right| + C} \hfill \\
Substituting\,\,u = {x^4} + 4{x^2} + 7\,\,for\,\,u\,\,gives \hfill \\
\frac{1}{4}\ln \,\left( {{x^4} + 4{x^2} + 7} \right) + C \hfill \\
\end{gathered} \]
