Answer
\[ = \frac{{\,{{\left( {{x^2} - 1} \right)}^{100}}}}{{100}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {2x\,{{\left( {{x^2} - 1} \right)}^{99}}} \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = {x^2} - 1\,\,\,\,then\,\,\,du = 2xdx \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
\int_{}^{} {2x\,{{\left( {{x^2} - 1} \right)}^{99}}\,\,} \,\, = \int_{}^{} {{u^{99}}\,du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \frac{{{u^{100}}}}{{100}} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,u = {x^2} - 1 \hfill \\
\hfill \\
= \frac{{\,{{\left( {{x^2} - 1} \right)}^{100}}}}{{100}} + C \hfill \\
\end{gathered} \]