Math 22 Exponential and Logarithmic Integrals
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Integrals of Exponential Functions
Let 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 u}
be a differentiable function of , then
Exercises 1 Find the indefinite integral
1)
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2)
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| Let , so , so |
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3)
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4)
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| Let 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 u=2x-5} , so 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 du=2dx} , so 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 dx=\frac{du}{2}} |
| 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 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 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 u}
be a differentiable function of 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 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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