Calculus
posted by Leanna .
I don't know how to do the integral of e^(lnx^2)dx and the integral of (sin sqrtx)/(sqrtx) dx

1. ∫e^(lnx^2)dx
use the identity e^(ln(y)) = y to simplify the expression.
2. try the substitution u=sqrt(x). 
Thanks for help on 1.
on 2. if i do u=sqrt(x) my du is 1/2x^(1/2) and that means my du is in the denominator. So it would read 2integral of sin(u)/du 
Nope, not on two.
You cant solve it that way easily.
This is difficult. Brake the sin function into its series equivalent, and integrate the series.
http://reference.wolfram.com/mathematica/ref/SinIntegral.html 
forget that last answer. I am tired.

For 2, almost, but not quite!
Start with:
u=√x
du = (1/2)dx/√x
dx/√x = 2du
so
∫sin(√x) dx/√x
=∫sin(u)*2du
=2cos(u)
=2cos(√x)