3

Find the integration enter image description here

my try :

enter image description here

Unable to solve further

cattt
  • 97
  • It may help to note that $$\int_0^1\frac{te^{2t}}{e(1-t)+te^{2t}}dt=\int_0^1\frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+e\int_0^1\frac{t-1}{e(1-t)+te^{2t}}dt\=1+e\int_0^1\frac{t-1}{e(1-t)+te^{2t}}dt$$ – clathratus Jan 27 '19 at 06:54

1 Answers1

2

Being almost blind, I have problems trying to read your notes.

What it seems to me is that there is a symmetry around $x=\frac 32$. So, let $x=y+\frac 32$ to make the integrand to be $$f(y)=\frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes $$f(y)+f(-y)=1$$ So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-\frac 12,0)$, $(\frac 12,0)$, $(\frac 12,1)$.

Now, a simple result.

Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !

Claude Leibovici
  • 260,315