How do I Integrate this u substitution with limits.

sg001
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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.
 
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Should it become this?
\int^{2}_{1}(1+u^{2})^{-1}du
= \left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}
 
th4450 said:
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).
 
Oops haha
should let u = tanθ
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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