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How do I Integrate this! u substitution with limits.

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sg001
#1
Feb21-12, 04:15 AM
P: 134
1. The problem statement, all variables and given/known data

Find ∫e^x/ (1+e^2x). dx , with limits ln 2 & 0
given u= e^x

2. Relevant equations



3. The attempt at a solution

u= e^x
du/dx = e^x
dx= du/e^x

sub limits of ln2 & 0 → u

Hence, limits 2 & 1

Therefore,

∫u* (1+e^2x)^-1* du/e^x

= ∫ u/ (u + e^3x)
= ∫ u/ e^3x
= ∫ 1/e^2x

= -e^-x
= -1/u
plugging in limits of 2 &1

Therefore, 0.2325...

Although i could not find this on the answer sheet did i do something wrong?
Please help, Thankyou.
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th4450
#2
Feb21-12, 06:06 AM
P: 38
Should it become this?
[itex]\int^{2}_{1}(1+u^{2})^{-1}du[/itex]
= [itex]\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}[/itex]
HallsofIvy
#3
Feb21-12, 06:16 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,345
Quote Quote by th4450 View Post
Should it become this?
[itex]\int^{2}_{1}(1+u^{2})^{-1}du[/itex]
Yes, this is correct. Letting [itex]u= e^x[/itex], [itex]du= e^xdx[/itex] so the numerator is just du and [itex](1+ e^{2x}[/itex] becomes [itex]1+ u^2[/itex]

= [itex]\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}[/itex]
However, this is incorrect. Yes, [itex]\int 1/u du= ln|u|+ C[/itex] but if you have a f(u) rather than u, you cannot just divide by f'(u)- that has to be already in the integral in order to make that substitution.

Instead, look up the derivative of arctan(u).

th4450
#4
Feb21-12, 09:06 AM
P: 38
How do I Integrate this! u substitution with limits.

Oops haha
should let u = tanθ


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