Answer
\[ - \frac{1}{{2\,{{\left( {{x^2} + x} \right)}^2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{2x + 1}}{{\,{{\left( {{x^2} + x} \right)}^3}}}dx} \hfill \\
Let\,\,u = {x^2} + x\,,\,\,So\,\,that \hfill \\
du = \,\left( {2x + 1} \right)dx \hfill \\
Then \hfill \\
\int_{}^{} {\frac{{2x + 1}}{{\,{{\left( {{x^2} + 1} \right)}^3}}}dx} = \int_{}^{} {\frac{{du}}{{{u^3}}}} \hfill \\
\int_{}^{} {{u^{ - 3}}du} \hfill \\
Power\,\,rule \hfill \\
\frac{{{u^{ - 3 + 1}}}}{{ - 3 + 1}} + C \hfill \\
- \frac{1}{{2{u^2}}} + C \hfill \\
Substituting\,\,u = {x^2} + x\,for\,\,u \hfill \\
- \frac{1}{{2\,{{\left( {{x^2} + x} \right)}^2}}} + C \hfill \\
\end{gathered} \]
