\(\int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C\)
\[\begin{eqnarray} \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C \\ \mu 为常数, 但\mu \ne 0 \\ \\ 推导过程如下: \\ (\frac{1}{\mu+1} x^{\mu+1})' =\frac{(x^{\mu+1})' \cdot \mu +1 - x^{\mu+1} \cdot (\mu+1)'}{(\mu+1)^{2}} \\ \\ \because (\mu+1)'=0, \enspace (x^{\mu+1})'=(\mu+1) x^{\mu+1-1} \\ \\ \Rightarrow \frac{(\mu+1) x^{\mu} \cdot \mu +1 - x^{\mu+1} \cdot 0}{(\mu+1)^{2}} \\ \\ \frac{((\mu+1) x^{\mu} \cdot \mu +1}{(\mu+1)^{2}}=x^{\mu} \\ \\ (\frac{1}{\mu+1} x^{\mu+1})'=x^{\mu} \\ \\ \therefore \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C \end{eqnarray} \]\(\int a^{x}dx=\frac{a^{x}}{\ln{a}}+C\)
\[ \begin{eqnarray} \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C, \enspace (a为常数,a>0,a \ne 1) \\ \\ 推导过程如下: \\ 已知: \enspace (a^{x})'=a^{x}\ln{a}, \enspace (\ln{a})'=0 \\ \\ \frac{(a^{x})'}{(\ln{a})'}=\frac{(a^{x})'\ln{a}-a^{x}(\ln{a})'}{\ln^{2}{a}} \\ \\ \frac{a^{x}\ln{a} \cdot \ln{a}-a^{x}\cdot 0}{\ln^{2}{a}} \\ \\ \Rightarrow \frac{a^{x}\ln{a} \cdot \ln{a}}{\ln^{2}{a}}=a^{x} \\ \\ \therefore \frac{a^{x}}{\ln{a}}之导为a^{x} \\ \\ \therefore \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C , \enspace (a>0,a \ne 1) \end{eqnarray} \]标签：基本,frac,cdot,积分,mu,ln,int,dx From： https://www.cnblogs.com/Preparing/p/18168542