Math 22 Exponential and Logarithmic Integrals

Integrals of Exponential Functions

 Let $u$ be a differentiable function of $x$, then
 $\int e^{x}dx=e^{x}+C$
 
 $\int e^{u}{\frac {du}{dx}}dx=\int e^{u}du=e^{u}+C$

Exercises 1 Find the indefinite integral

1) $\int 3e^{x}dx$

Solution:
$\int 3e^{x}dx=3\int e^{x}=3e^{x}+C$

2) $\int 3e^{3x}dx$

Solution:
Let $u=3x$, so $du=3dx$, so $dx={\frac {du}{3}}$
Consider $\int 3e^{3x}dx=\int 3e^{u}{\frac {du}{3}}=\int e^{u}du=e^{u}+C=e^{3x}+C$

3) $\int (3e^{x}-6x)dx$

Solution:
$\int (3e^{x}-6x)dx=\int (3e^{x})dx-\int 6xdx=3e^{x}-3x^{2}+C$

4) $\int e^{2x-5}dx$

Solution:
Let $u=2x-5$, so $du=2dx$, so $dx={\frac {du}{2}}$
Consider $\int e^{2x-5}dx=\int e^{u}{\frac {du}{2}}={\frac {1}{2}}\int e^{u}du={\frac {1}{2}}e^{u}+C={\frac {1}{2}}e^{2x-5}+C$

Using the Log Rule

 Let $u$ be a differentiable function of $x$, then
 \int\frac{1}{x}=\ln\abs{x}+C
 
 \int\frac{1}{u}\frac{du}{dx}dx=\int\frac{1}{u}du=\ln\abc{u}+C

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