# Yet another integration problem.

• Feb 26th 2008, 11:39 AM
quarks
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:

$\displaystyle \ \int_1^S \frac{k_1}{k_2-n} \ dn$
• Feb 26th 2008, 11:50 AM
Jhevon
Quote:

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:

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

$\displaystyle k_1 \mbox{ and }k_2$ are constants.

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

remember, treat $\displaystyle k_1 \mbox{ and } k_2$ as if they were the numbers you love
• Feb 26th 2008, 11:57 AM
quarks
So, then:
$\displaystyle \int_1^S \frac{k_1}{k_2-n} \ dn$

$\displaystyle u=k_2-n$

$\displaystyle du=-dn$

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

$\displaystyle -k_1 \ln\frac{1}{k_2-n}$, evaluated from 1 to S.
• Feb 26th 2008, 12:11 PM
Jhevon
Quote:

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

$\displaystyle u=k_2-n$

$\displaystyle du=-dn$

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

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

no. it's $\displaystyle -k_1 \ln \left| k_2 - n\right|$

and the integral is with respect to u