- #1
Gwozdzilla
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Homework Statement
We know that F(x) = [itex]\int^{x}_{0}e^{e^{t}} dt[/itex] is a continuous function by FTC1, though it is not an elementary function. The Functions [itex]\int\frac{e^{x}}{x}dx[/itex] and [itex]\int\frac{1}{lnx}dx[/itex] are not elementary funtions either but they can be expressed in terms of F.
a) [itex]\int^{2}_{1}\frac{e^{x}}{x}dx[/itex]
b)[itex]\int^{3}_{2}\frac{1}{lnx}dx[/itex]
Homework Equations
I only used and u-substitution in my attempt at a solution, but I suppose integration by parts and trigonometric substitution is fair game as well.
Integration by parts: [itex]\int udv[/itex] = uv - [itex]\int vdu[/itex]
The Attempt at a Solution
a) [itex]\int^{2}_{1}\frac{e^{x}}{x}dx[/itex]
Let x = et
dx =etdx
[itex]\int\frac{e^{e^{t}}e^{t}}{e^{t}}dt[/itex]
[itex]\int^{2}_{1}e^{e^{t}} dt[/itex]
Is this answer correct? If it isn't, where did I go wrong? If it is, how do I put it in terms of F(x)?
b)[itex]\int^{3}_{2}\frac{1}{lnx}dx[/itex]
let u = ln(x)
du = 1/x dx
xdu = dx
eudu = dx
[itex]\int\frac{e^{u}}{u}dx[/itex]
Which this is just part a again.. but am I allowed to do u substitution that way by plugging in u into my du line since it ended up being in terms of an x I didn't have to substitute with?