5.5 The Substitution Rule/63

$\int _{0}^{a}x*{\sqrt {x^{2}+a^{2}}}dx$

${\begin{aligned}u&=x^{2}+a^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}$

${\begin{aligned}{\frac {1}{2}}\int _{a^{2}}^{2a^{2}}{\sqrt {u}}\,\cdot du={\frac {1}{2}}\,\cdot {\frac {a^{{\frac {1}{2}}+1}}{1/2+1}}{\bigg |}_{2a^{2}}^{a^{2}}\end{aligned}}$