Hey 99.95.
Use the hint that e^x*e^x = e^2x where du = e^xdx and u = e^x.
imgur: the simple image sharer
for the first one,
i tried:
let e^x = u
du/dx = e^x
but then i couldn't put that back in (du = e^x dx is of no use)
then I tried let
e^2x = u
du = 2e^2x dx
putting it back in i had
1/2 Integral of [ ucos(u^1/2)du ]
let f = u let dg/du = cos(u^1/2)
df/du = 1 g = ?? Got ugly here i think.
Any ideas?
Try introducing a SQRT(x)*SQRT(x) term and just check that this is OK for the integral (it should be, but you'll need to check for yourself just in case). So instead of finding e^(SQRT(x))dx find SQRT(x)e^(SQRT(x))/SQRT(x)dx (Also this should work because you only have a single point where this breaks down which means it should be OK).