INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 0 answersPost by Arshin Jain on April 25, 2014This 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.