INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 1 answerLast reply by: Richard GregoryThu Nov 28, 2013 1:44 AMPost by Joel Fredin on November 12, 2013In your last example, i don't really understand how you could multiply with (2x-1) where did you got your "-1" from? :) Other than that, your making really good tutorials, keep it up!

### Integrals of Exponential Functions

• All other integration rules still apply
• Look for lots of substitution or chain rules applied with exponential functions

### Integrals of Exponential Functions

∫ex dx
∫ex dx = ex + c
∫e5x dx
• u = 5x
• du = 5  dx
• ∫e5x dx = [1/5] ∫eu du
• ∫e5x dx = [1/5] eu + c
∫e5x dx = [1/5] e5x + c
∫x ex2 dx
• u = x2
• du = 2x  dx
• ∫x ex2 dx = ∫ex2 [1/2] 2x  dx
• ∫x ex2 dx = [1/2] ∫eu du
• ∫x ex2 dx = [1/2] eu + c
∫x ex2 dx = [1/2] ex2 + c
∫9x dx
• ∫9x dx = ∫e(ln9)  x
• u = ln9  x
• du = ln9  dx
• ∫9x dx = [1/ln9] ∫eln9  x ln9  dx
• ∫9x dx = [1/ln9] ∫eu du
• ∫9x dx = [(eu)/ln9] + c
• ∫9x dx = [(eln9, x)/ln9] + c
∫9x dx = [(9x)/ln9] + c
∫57x dx
• u = 7x
• du = 7  dx
• ∫57x dx = [1/7] ∫57x 7  dx
• ∫57x dx = [1/7] ∫5u du
• ∫57x dx = [1/7] [(5u)/ln5] + c
∫57x dx = [(57x)/7 ln5] + c
∫e2x + 1 dx
• u = 2x + 1
• du = 2  dx
• ∫e2x + 1 dx = [1/2] ∫e2x + 1 2  dx
• ∫e2x + 1 dx = [1/2] ∫eu du
• ∫e2x + 1 dx = [1/2] eu + c
∫e2x + 1 dx = [(e2x + 1)/2] + c
∫cos(x)esinx dx
• u = sinx
• du = cosx  dx
• ∫cos(x)esinx dx = ∫esinx cosx  dx
• ∫cos(x)esinx dx = ∫eu du
• ∫cos(x)esinx dx = eu + c
∫cos(x)esinx dx = esinx + c
∫[(e√x)/(√x)] dx
• u = √x
• du = [1/(2√x)] dx
• ∫[(e√x)/(√x)] dx = 2 ∫e√x [1/(2√x)] dx
• ∫[(e√x)/(√x)] dx = 2 ∫eu du
• ∫[(e√x)/(√x)] dx = 2 eu + c
∫[(e√x)/(√x)] dx = 2 e√x + c
∫[(e√{7x})/(√x)] dx
• u = √{7x}
• du = [1/2] [1/(√{7x})] [d/dx] 7x
• du = [7/2] [1/(√{7x})] dx
• du = [7/2] [1/((√7)√x)] dx
• du = [(√7)/2] [1/(√x)] dx
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] ∫e√{7x} [(√7)/(2√x)] dx
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] ∫eu du
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] eu + c
∫[(e√{7x})/(√x)] dx = [2/(√7)] e√{7x} + c
∫sec2(5x) 4tan5x dx
• u = tan5x
• du = 5 sec2 (5x)  dx
• ∫sec2(5x) 4tan5x dx = [1/5] ∫4tan5x 5 sec2 (5x)  dx
• ∫sec2(5x) 4tan5x dx = [1/5] ∫4u du
• ∫sec2(5x) 4tan5x dx = [1/5] [(4u)/ln4] + c
∫sec2(5x) 4tan5x dx = [(4tan5x)/5 ln4] + c

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Integrals of Exponential Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Types of Integrals: Exponential Functions 0:09
• Rule 1
• Rule 2
• Example 1
• Example 2 2:54
• Example 3 4:19
• Example 4 5:19
• Example 5 7:37
• Example 6 9:04