# Thread: Yet another integration problem.

1. ## Yet another integration problem.

It tells me to use substitution, but I have the problem (like many other people, probably), of being totally lost once actual expressions and numbers are removed. So, can anyone give me some help with this integral, using the method of substitution:

$\
\int_1^S \frac{k_1}{k_2-n} \ dn
$

2. Originally Posted by quarks
It tells me to use substitution, but I have the problem (like many other people, probably), of being totally lost once actual expressions and numbers are removed. So, can anyone give me some help with this integral, using the method of substitution:

$\
\int_1^S \frac{k_1}{k_2-n} \ dn
$
$k_1 \mbox{ and }k_2$ are constants.

make the substitution $u = k_2 - n$ and continue

remember, treat $k_1 \mbox{ and } k_2$ as if they were the numbers you love

3. So, then:
$\int_1^S \frac{k_1}{k_2-n} \ dn$

$u=k_2-n$

$du=-dn$

$-k_1$ $\int_1^S \frac{1}{u} \ dn$

$-k_1 \ln\frac{1}{k_2-n}$, evaluated from 1 to S.

4. Originally Posted by quarks
So, then:
$\int_1^S \frac{k_1}{k_2-n} \ dn$

$u=k_2-n$

$du=-dn$

$-k_1$ $\int_1^S \frac{1}{u} \ dn$

$-k_1 \ln\frac{1}{k_2-n}$, evaluated from 1 to S.
no. it's $-k_1 \ln \left| k_2 - n\right|$

and the integral is with respect to u