# Math 22 Exponential and Logarithmic Integrals

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## Integrals of Exponential Functions

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

${\displaystyle \int e^{u}{\frac {du}{dx}}dx=\int e^{u}du=e^{u}+C}$


Exercises 1 Find the indefinite integral

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

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

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

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

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

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

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

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

## Using the Log Rule

 Let ${\displaystyle u}$ be a differentiable function of ${\displaystyle x}$, then
${\displaystyle \int {\frac {1}{x}}=\ln |x|+C}$

${\displaystyle \int {\frac {1}{u}}{\frac {du}{dx}}dx=\int {\frac {1}{u}}du=\ln |u|+C}$


Exercises 2 Find the indefinite integral

1) ${\displaystyle \int {\frac {3}{x}}dx}$

Solution:
${\displaystyle \int {\frac {3}{x}}dx=3\int {\frac {1}{x}}=3\ln |x|+C}$

2) ${\displaystyle \int {\frac {3x}{x^{2}}}dx}$

Solution:
Let ${\displaystyle u=x^{2}}$, so ${\displaystyle du=2xdx}$, so ${\displaystyle dx={\frac {du}{2x}}}$
Consider ${\displaystyle \int {\frac {3x}{x^{2}}}dx=\int {\frac {3x}{u}}{\frac {du}{2x}}=\int {\frac {3}{2}}{\frac {1}{u}}du={\frac {3}{2}}\int {\frac {1}{u}}du={\frac {3}{2}}\ln |u|+C={\frac {3}{2}}\ln |x^{2}|+C}$

3) ${\displaystyle \int {\frac {3}{3x+5}}dx}$

Solution:
Let ${\displaystyle u=3x+5}$, so ${\displaystyle du=2dx}$, so ${\displaystyle dx={\frac {du}{3}}}$
Consider ${\displaystyle \int {\frac {3}{3x+5}}dx=\int {\frac {3}{u}}{\frac {du}{3}}=\int {\frac {3}{3}}{\frac {1}{u}}du=\int {\frac {1}{u}}du=\ln |u|+C=\ln |3x+5|+C}$