Thread: simple intergration help

ok
what is the integral of ( 1/2t)

i did a u sub

so u = 2t
du = 2

2 * ln2t

is this correct?

which is a standard integral.

Originally Posted by o_O

which is a standard integral.

ah so it is lnt/2

?

yeah it is.. ty sir.

No. You will get 1/2 ln(t).
They are not the same.
1/2 ln(t) <> ln(t/2). They maybe different.

Originally Posted by oswaldo

No. You will get 1/2 ln(t).
They are not the same.
1/2 ln(t) <> ln(t/2). They maybe different.

ok but what am i doing wrong then


int (1/2t)

so if u just do a u sub instead of pulling the 1/2 out.

u = 2t
du = 2du

(1/2) int [(2du)/(2t)

=

1/2 lnu

1/2 * ln 2t


why is it 1/2 * lnt and not 1/2 * ln2t

what did i do wrong?

You don't need any kind of u or whatever conversion. It is a simple basic integral. All you need to know is:
int(1/t) = ln(t). THAT'S IT.

1/2 is just a constant. So the answer of your questions is:
int(1/2t)=(1/2) int(1/t)=(1/2) ln(t). AND THAT'S IT.
Which may be different than ln(t/2).

a*ln(t)=ln(a*t) is NOT always true.

So (1/2)*ln(t) = ln((1/2)*t) is NOT always true.

-O

If for some reason you decided to use a u-substitition:



We then get:

Looks a bit different. But recall:

So:

But is nothing more than a constant. So we'll combine it with to make a new constant of integration which we'll call .

i.e.

In other words, and are two legit antiderivatives of