Integrating log

nobodythere

Hello,
I'm trying to integrate

ln(1/(1-(6t/101))dt

but can't get the right answer out.

I do:

x= 1-(6t/101)
dx = -6/101 dt

giving me

(-6/101)ln(1/x)dx

I then integrate by parts:

u=(-6/101)ln(1/x)
du=(-6/101)(-1/x)dx

dv=dx
v=x

uv - Integration(vdu) =
(-6/101)(x)ln(1/x) - (-6/101)Integration(-x/x dx)
= (-6/101) (xln(1/x) + x)

well i'll stop there because I don't get the right answer out. Help please

Michael Collins

Your answer looks perfectly fine to me, as a quick differentiation of your last line should tell you!

nobodythere

The answer this gives me is different: http://integrals.wolfram.com/index.jsp

It's part of a Maths Physics question and I've to show that when evaluating the integral for t between 13.3 and 0 I should get about 8.99

Using my method (after subbing t back in) I get an oddball answer (less than 1) but the integral that that integrator gives me turns out right

Michael Collins

The function you have will give you about 7.784 if you integrate it between 0 and 13.3. Are you sure you have the original function right? There's nothing wrong with your maths...

And you also get the same answer from the Mathematica tool you linked to in your last post

MathsManiac

I think you've the initial fraction the wrong way up when you've transformed from t to x. Shouldn't it be
int[(-101/6)ln(1/x)dx]?

Does this sort it out?

(p.s., I also got -7.784045016, like Michael C, going from 13.3 to 0.)

Michael Collins

MathsManiac wrote: »

I think you've the initial fraction the wrong way up when you've transformed from t to x. Shouldn't it be
int[(-101/6)ln(1/x)dx]?

Ah yes, that's it! I somehow overlooked that part...

MathsManiac

Also just spotted what I think is a slightly easier way.

The initial function can be simplified to ln(101/(101-6t)).
Break this up using the log rule:
ln(101)-ln(101-6t)
Integrating the first bit is just a constant, and you can sidestep the int by parts for the second bit if you know the standard result that int(ln(x)) is xln(x)-x. (You still need a subs, but it's linear so you can handle it by just chaining out the -1/6.)

p.s. still 7.784, by the way!

nobodythere

Thanks for your replies, 7.784 is the answer I'm looking for.

Had a few looks over it and can't understand why my fraction would be 101/6 and not 6/101

EDIT: Never mind, got it, thanks for your help!