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

Exercises 2 Find the indefinite integral

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

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

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

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

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

Solution:
Let $u=3x+5$, so $du=2dx$, so $dx={\frac {du}{3}}$
Consider Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}

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