can u help me w/this ...integrals stuff...?

Question

the part where i circled...after the u-substitution and find out what their value is, what do u do next? can u plz guide me how to get those steps...

http://i181.photobucket.com/albums/x85/ctti_3/Untitled-83.jpg

Answer 1

If I understand you right, you're asking how you get from the determination of the u substitution values to the rewritten equation in the circle?

First, the "du" should include a 'dx' at the end of it:
du = 2e^x dx

So then you have
I = igrl e^x / sqrt(2e^x - 1) dx
u = 2e^x - 1
du = 2e^x dx

You substitute u for the 2e^x - 1 in the square root:

I = igrl e^x / sqrt(u) dx

(dx is actually part of the numerator here, so I'll just rewrite that)

I = igrl e^x dx / sqrt(u)

du = 2e^x dx
du/2 = e^x dx

So substitute du/2 for your numerator:

I = igrl du/2 / sqrt(u)

The 1/2 is a constant, so move it outside the integral:

I = 1/2 igrl 1/sqrt(u) du

Hope that helps.

Answer 2

The part you circled is perfectly correct.
Just substitute for dx and √(2e^x-1).
If you put 1/2 outside the integral and 2 inside
you get du/√u exactly.

Answer 3

I got sqrt(3) exactly. You need to take the u^(3/2) out of denominator, will be u^-(3/2), which simplifies to u * sqrt u.
Plug your value back in for u and simplify and apply the bounds (0 to ln 2). Everything should become 0, and you are left with sqrt 3.