# Math Help - improper integration

1. ## improper integration

I am trying to do this
integral sign going from 0 to 4 (0 on the bottom, 4 on the top) 1/(4-x) dx
i got stuck.
Thanks a lot!!!

2. $\int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$

3. Originally Posted by Krizalid
$\int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$
Yeah i was able to get that far, but not so sure how to solve it.

4. Originally Posted by davecs77
Yeah i was able to get that far, but not so sure how to solve it.
If you got that far, what's the problem?

$\lim_{a \to 4} \int_0^a \frac {1}{4 - x}~dx$

$= \lim_{a \to 4} \left[ - \ln (4 - x) \right]_{0}^{a}$

$= \lim_{a \to 4} \left( - \ln (4 - a) + \ln 4 \right)$

Now take the limit