Question

integrate du/3u

and integrate

x/x^2 dx

it might be simple...i just need a head start...tnx

Answer 1

do i use the subsitution rule?

Answer 2

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

Answer 3

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