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Math Help - Yet another integration problem.

  1. #1
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    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:

    \<br />
\int_1^S \frac{k_1}{k_2-n}  \ dn<br />
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by quarks View Post
    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:

    \<br />
\int_1^S \frac{k_1}{k_2-n}  \ dn<br />
    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
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  3. #3
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    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.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by quarks View Post
    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
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