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

Answer 1

1. ∫e^(lnx^2)dx
use the identity e^(ln(y)) = y to simplify the expression.
2. try the substitution u=sqrt(x).

Answer 2

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

Answer 3

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

Answer 4

forget that last answer. I am tired.

Answer 5

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)