If we choose $u=46+e^x,$ then $du=e^{x}dx.$

If $u=46+e^x$ is substituted into ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{46+e^x}dx,$ then we have

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{46+e^x}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^x\sqrt[\phantom{2}]{u}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{u}(e^{x}dx)$

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

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