1. ## Improper Integration

The integral from 1 to infinity [sorry i don't know how to put it in LaTex ], of:
$\frac {3dx}{3x-x^2}$

From what My teacher showed me, you can use Ln, or partial fractions to solve this, but I couldn't understand either method, how do you do it?

2. Originally Posted by >_<SHY_GUY>_<
The integral from 1 to infinity [sorry i don't know how to put it in LaTex ], of:
$\frac {3dx}{3x-x^2}$

From what My teacher showed me, you can use Ln, or partial fractions to solve this, but I couldn't understand either method, how do you do it?
Resolve into partial fractions:

$\frac{3}{3x-x^2}=\frac{A}{x}+\frac{B}{3-x}$

and a quick calculation shows that:

$\frac{3}{3x-x^2}=\frac{1}{x}+\frac{1}{3-x}$

Which has resolved the integrand into the sum of two terms both of which are easily integrable (in terms of $\ln$ )

CB

3. Originally Posted by CaptainBlack
Resolve into partial fractions:

$\frac{3}{3x-x^2}=\frac{A}{x}+\frac{B}{3-x}$

and a quick calculation shows that:

$\frac{3}{3x-x^2}=\frac{1}{x}+\frac{1}{3-x}$

Which has resolved the integrand into the sum of two terms both of which are easily integrable (in terms of $\ln$ )

CB
from there you solve for A and B and then take the integrals of those fractions...Im on the right track right?

4. Yes. In the second step CB solved for A, B and found A=B=1. Now you can integrate both fractions separately, which is easy.

5. Originally Posted by >_<SHY_GUY>_<
from there you solve for A and B and then take the integrals of those fractions...Im on the right track right?
yes,..go on