\(\int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C\)
\[\begin{eqnarray}
\int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C
\\
\mu 為常數, 但\mu \ne 0
\\ \\
推導過程如下: \\
(\frac{1}{\mu+1} x^{\mu+1})'
=\frac{(x^{\mu+1})' \cdot \mu +1 - x^{\mu+1} \cdot (\mu+1)'}{(\mu+1)^{2}}
\\ \\
\because (\mu+1)'=0, \enspace (x^{\mu+1})'=(\mu+1) x^{\mu+1-1}
\\ \\
\Rightarrow \frac{(\mu+1) x^{\mu} \cdot \mu +1 - x^{\mu+1} \cdot 0}{(\mu+1)^{2}}
\\ \\
\frac{((\mu+1) x^{\mu} \cdot \mu +1}{(\mu+1)^{2}}=x^{\mu}
\\ \\
(\frac{1}{\mu+1} x^{\mu+1})'=x^{\mu}
\\ \\
\therefore \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C
\end{eqnarray}
\]
\(\int a^{x}dx=\frac{a^{x}}{\ln{a}}+C\)
\[
\begin{eqnarray}
\int a^{x}dx=\frac{a^{x}}{\ln{a}}+C, \enspace (a為常數,a>0,a \ne 1)
\\ \\
推導過程如下: \\
已知: \enspace (a^{x})'=a^{x}\ln{a}, \enspace
(\ln{a})'=0
\\ \\
\frac{(a^{x})'}{(\ln{a})'}=\frac{(a^{x})'\ln{a}-a^{x}(\ln{a})'}{\ln^{2}{a}}
\\ \\
\frac{a^{x}\ln{a} \cdot \ln{a}-a^{x}\cdot 0}{\ln^{2}{a}}
\\ \\
\Rightarrow \frac{a^{x}\ln{a} \cdot \ln{a}}{\ln^{2}{a}}=a^{x}
\\ \\
\therefore \frac{a^{x}}{\ln{a}}之導為a^{x}
\\ \\
\therefore \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C
, \enspace (a>0,a \ne 1)
\end{eqnarray}
\]