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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (1)

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Post by Arshin Jain on April 25, 2014

This certainly was delightful!

{Especially the last problem :D }

Integrals of Natural Logarithmic Functions

    • Try to transform expression into a fraction with only a constant in the numerator
    • Substitute denominator with u
  • All other integration rules still apply

Integrals of Natural Logarithmic Functions

∫[1/x]  dx
∫[1/x]  dx = lnx + c
∫[ln12x/x]  dx
  • u = ln12x
  • du = [1/x] dx
  • ∫[ln12x/x]  dx = ∫ln12x [1/x] dx
  • ∫[ln12x/x]  dx = ∫u  du
  • ∫[ln12x/x]  dx = [(u2)/2] + c
∫[ln12x/x]  dx = [((ln12x)2)/2] + c
∫[1/x lnx] dx
  • u = lnx
  • du = [1/x] dx
  • ∫[1/x lnx] dx = ∫[1/lnx] [1/x] dx
  • ∫[1/x lnx] dx = ∫[1/u] du
  • ∫[1/x lnx] dx = lnu + c
∫[1/x lnx] dx = ln(lnx) + c
∫[(2x + 1)/(x2 + x +7)] dx
  • u = x2 + x + 7
  • du = (2x + 1) dx
  • ∫[(2x + 1)/(x2 + x +7)] dx = ∫[1/(x2 + x + 7)] (2x + 1) dx
  • ∫[(2x + 1)/(x2 + x +7)] dx = ∫[1/u] du
  • ∫[(2x + 1)/(x2 + x +7)] dx = lnu + c
∫[(2x + 1)/(x2 + x +7)] dx = ln(x2 + x + 7) + c
∫cotx sec2 x  dx
  • ∫cotx sec2 x  dx = ∫[(sec2 x)/tanx] dx
  • u = tanx
  • du = sec2 x  dx
  • ∫cotx sec2 x  dx = ∫[1/tanx] sec2 x  dx
  • ∫cotx sec2 x  dx = ∫[1/u] du
  • ∫cotx sec2 x  dx = lnu + c
∫cotx sec2 x  dx = ln(tanx) + c
∫[(e2x + 1)/(e2x + 2x)] dx
  • u = e2x + 2x
  • du = 2 e2x + 2
  • ∫[(e2x + 1)/(e2x + x)] dx = [1/2] ∫[1/(e2x + x)] 2(e2x + 1) dx
  • ∫[(e2x + 1)/(e2x + x)] dx = [1/2] ∫[1/u] du
  • ∫[(e2x + 1)/(e2x + x)] dx = [1/2] lnu + c
∫[(e2x + 1)/(e2x + x)] dx = [1/2] ln(e2x + 2x) + c
∫tan(3x) dx
  • ∫tan(3x) dx = ∫[sin(3x)/cos(3x)] dx
  • u = cos(3x)
  • du = −sin(3x) dx
  • ∫tan(3x) dx = − ∫[1/cos(3x)] (−sin(3x)) dx
  • ∫tan(3x) dx = − ∫[1/u] du
  • ∫tan(3x) dx = −lnu + c
∫tan(3x) dx = −ln(cos(3x)) + c
∫x cot(x2) dx
  • u = x2
  • du = 2x  dx
  • ∫x cot(x2) dx = [1/2] ∫cotu  du
  • ∫x cot(x2) dx = [1/2] ∫[cosu/sinu] du
  • Another substitution is needed
  • v = sinu
  • dv = cosu  du
  • ∫x cot(x2) dx = [1/2] ∫[1/v] dv
  • ∫x cot(x2) dx = [1/2] lnv + c
  • ∫x cot(x2) dx = [1/2] ln(sinu) + c
∫x cot(x2) dx = [1/2] ln(sinx2) + c
∫[(e2x)/((e2x + ex))] dx
  • ∫[(e2x)/((e2x + ex))] dx = ∫[(e2x)/(ex(ex + 1))] dx
  • ∫[(e2x)/((e2x + ex))] dx = ∫[(ex)/(ex + 1)] dx
  • u = ex + 1
  • du = ex dx
  • ∫[(e2x)/((e2x + ex))] dx = ∫[1/u] du
  • ∫[(e2x)/((e2x + ex))] dx = lnu + c
∫[(e2x)/((e2x + ex))] dx = ln(ex + 1) + c
∫[1/(sin−1 (x) √{1 − x2})] dx
  • u = sin−1 x
  • du = [1/(√{1 − x2})] dx
  • ∫[1/(sin−1 (x) √{1 − x2})] dx = ∫[1/u] du
  • ∫[1/(sin−1 (x) √{1 − x2})] dx = lnu + c
∫[1/(sin−1 (x) √{1 − x2})] dx = ln(sin−1 x) + 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.

Answer

Integrals of Natural Logarithmic 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: Natural Log Functions 0:09
    • Example 1
  • Example 2 2:06
  • Example 3 4:01
  • Example 4 5:37
  • Example 5 7:30
  • Example 6 9:05