If we choose $u=16+e^x,$ then $du=e^{x}dx.$

If $u=16+e^x$ is substituted into ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{16+e^x}dx,$ then we have

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{16+e^x}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{u}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{u}(e^{x}dx)$

We must also convert $e^{x}dx$ into an expression involving $u,$ but we already know that $e^{x}dx=du.$