Question

\[\int\limits\limits\limits_{0}^{2}e^{\sqrt{2x}}dx\]

Answer 1

The problem was:\[\int\limits\limits\limits_{0}^{2}e^{\sqrt{2x}}dx\]I started with integration by parts:\[u = e^{\sqrt{2x}}\]\[dv=dx\]\[du = \frac{e^{\sqrt{2x}}}{\sqrt{2x}}dx\]\[v=x\]So...\[=xe^{\sqrt{2x}}-\int\limits \frac{xe^{\sqrt{2x}}}{\sqrt{2x}}dx\]Then a 2nd integration by parts:\[u=x\]\[dv = \frac{e^{\sqrt{2x}}}{\sqrt{2x}}dx\]\[du = dx\]\[v = e^{\sqrt{2x}}\]So...\[\int\limits\limits e^{\sqrt{2x}}dx = xe^{\sqrt{2x}} - xe^{\sqrt{2x}}+ \int\limits\limits e^{\sqrt{2x}}dx\]Which gives me\[\int\limits\limits e^{\sqrt{2x}}dx = \int\limits\limits e^{\sqrt{2x}}dx\]So I'm back where I started....what do you do?

Answer 2

First of all: If you ise integration by parts then in the second integra it is pointless to do what you did you will always end up with the original integral.
To do this first put x=t^2. dx=2tdt
so it becomes integral of 2te^sqrt(2)t now integrate this by parts.