James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
Hi all, I am a bit stuck on this problem and would appreciate some pointers. Here is the problem and how I have attempted it so far.
"Show that
[math]\int_{\ln4}^{\ln3} \frac{e^{4x}}{e^x - 2} = \frac{a}{b} +c \ln d[/math]
where a, b, c and d are integers to be found. Use
[math]u^2 = e^x - 2[/math] "
To evaluate the integral and prove the statement, find the integral, which I have tried to do like this:
[math]2udu = e^x dx[/math]
and
[math]e^x = u^2 + 2[/math]
so
[math]dx = \frac{2udu}{u^2+2}[/math]
now
[math]\int \frac{e^{4x}}{e^x - 2}dx = \frac{e^{4x}}{u^2} \frac{2udu}{u^2+2}[/math]
[math]e^x = u^2 +2[/math]
[math]e^{4x} = (e^x)^4 = (u^2 + 2)^4[/math]
so
[math]\int \frac{2u(u^2 +2)^4}{u^2(u^2+2)}du[/math]
cancelling,
[math]\int \frac {2(u^2 +2)^3}{u}du[/math]
So the substitution has not really led to a simplified expression and I do not know how to continue from here. Is there something better I could have done?
"Show that
[math]\int_{\ln4}^{\ln3} \frac{e^{4x}}{e^x - 2} = \frac{a}{b} +c \ln d[/math]
where a, b, c and d are integers to be found. Use
[math]u^2 = e^x - 2[/math] "
To evaluate the integral and prove the statement, find the integral, which I have tried to do like this:
[math]2udu = e^x dx[/math]
and
[math]e^x = u^2 + 2[/math]
so
[math]dx = \frac{2udu}{u^2+2}[/math]
now
[math]\int \frac{e^{4x}}{e^x - 2}dx = \frac{e^{4x}}{u^2} \frac{2udu}{u^2+2}[/math]
[math]e^x = u^2 +2[/math]
[math]e^{4x} = (e^x)^4 = (u^2 + 2)^4[/math]
so
[math]\int \frac{2u(u^2 +2)^4}{u^2(u^2+2)}du[/math]
cancelling,
[math]\int \frac {2(u^2 +2)^3}{u}du[/math]
So the substitution has not really led to a simplified expression and I do not know how to continue from here. Is there something better I could have done?