How do I Integrate this u substitution with limits.

[sg001]

Homework Statement

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

Homework Equations

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.

 

[th4450]

Should it become this?
$\int^{2}_{1}(1+u^{2})^{-1}du$
= $\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}$

 

[HallsofIvy]

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).

 

[th4450]

Oops haha
should let u = tanθ