integrate
posted by allison .
integrate du/3u
and integrate
x/x^2 dx
it might be simple...i just need a head start...tnx

do i use the subsitution rule?

to be more specific...i know its best for me to attempt the problem
so for the first problem i used substitution as
v=3u
dv=3du
du=dv/3
so answer i got was 1/3 ln3u
is that correct?
and on the second on i tried
x*x^2 as x^3 dx so
ans is 1/2x^2
is that correct? please let me know before i go ahead...i have a test tommorow and i am trying to do as many practice problem as i can...tnx 
du/3u = (1/3) du/u
which is (1/3)[ ln u + ln constant ]
> (1/3) ln ( c u)
c = 3 works
so
(1/3) ln 3u + constant
x/x^2 = 1/ x
so
dx/x > ln x + constant
I think x^1*x^2 = x^1 = 1/x