# How do I Integrate this! u substitution with limits.

 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.
 P: 38 Should it become this? $\int^{2}_{1}(1+u^{2})^{-1}du$ = $\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}$
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P: 39,301
 Quote by th4450 Should it become this? $\int^{2}_{1}(1+u^{2})^{-1}du$
Yes, this is correct. Letting $u= e^x$, $du= e^xdx$ so the numerator is just du and $(1+ e^{2x}$ becomes $1+ u^2$

 = $\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}$
However, this is incorrect. Yes, $\int 1/u du= ln|u|+ C$ 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).

 P: 38 How do I Integrate this! u substitution with limits. Oops haha should let u = tanθ

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