Math 22 Exponential and Logarithmic Integrals - Revision history

Tphan046: /* Using the Log Rule */

2020-08-15T16:08:27Z

Using the Log Rule

2020-08-15T16:07:55Z

2020-08-15T15:53:59Z

2020-08-15T15:51:29Z

2020-08-15T15:43:16Z

Using the Log Rule

2020-08-15T15:42:48Z

Using the Log Rule

2020-08-15T15:42:12Z

2020-08-15T15:39:43Z

2020-08-15T15:39:22Z

Integrals of Exponential Functions

2020-08-15T15:39:13Z

@@ Line 69: / Line 69: @@
 |Let <math>u=3x+5</math>, so <math>du=2dx</math>, so <math>dx=\frac{du}{3}</math>
 |-
-|Consider <math>\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</math>
+|Consider <math>\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</math>
 |}

@@ Line 63: / Line 63: @@
 |}
-'''3)''' <math>\int (3e^x-6x)dx</math>
+'''3)''' <math>\int\frac{3}{3x+5}dx</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|<math>\int (3e^x-6x)dx=\int (3e^x)dx -\int 6xdx=3e^x-3x^2+C</math>
+|Let <math>u=3x+5</math>, so <math>du=2dx</math>, so <math>dx=\frac{du}{3}</math>
 |-
-|Consider <math>\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</math>
+|Consider <math>\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</math>
 |}

@@ Line 60: / Line 60: @@
 |Let <math>u=x^2</math>, so <math>du=2xdx</math>, so <math>dx=\frac{du}{2x}</math>
 |-
-|Consider <math>\int \frac{3x}{x^2}dx=\int\frac{3x}{u}\frac{du}{2x}=\int\frac{3}{2}\frac{1}{u}du=\frac{3}{2}\frac{1}{u}du=\frac{3}{2}\ln|u|+C=\frac{3}{2}\ln |x^2|+C</math>
+|Consider <math>\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</math>
 |}
@@ Line 76: / Line 76: @@
 |Let <math>u=2x-5</math>, so <math>du=2dx</math>, so <math>dx=\frac{du}{2}</math>
 |-
-|Consider <math>\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</math>
+|Consider <math>\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</math>
 |}

@@ Line 44: / Line 44: @@
    <math>\int\frac{1}{u}\frac{du}{dx}dx=\int\frac{1}{u}du=\ln|u|+C</math>
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 41: / Line 41: @@
 ==Using the Log Rule==
    Let <math>u</math> be a differentiable function of <math>x</math>, then
-   <math>\int\frac{1}{x}=\ln\abs{x}+C</math>
+   <math>\int\frac{1}{x}=\ln|x|+C</math>
-   <math>\int\frac{1}{u}\frac{du}{dx}dx=\int\frac{1}{u}du=\ln\abc{u}+C</math>
+   <math>\int\frac{1}{u}\frac{du}{dx}dx=\int\frac{1}{u}du=\ln|u|+C</math>
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 30: / Line 30: @@
 |}
-'''1)''' <math>\int e^{2x-5}dx</math>
+'''4)''' <math>\int e^{2x-5}dx</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;