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Math Help - Simple antiderivative question...

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    Simple antiderivative question...

    How do you take the antiderivative of something to the -1 power? Specifically I'm working with (12-10x)^-1. Using normal antideriv rules it would just be (12-10x)^0/0, which is irrational, could someone answer this for me? Thanks
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    Quote Originally Posted by fattydq View Post
    How do you take the antiderivative of something to the -1 power? Specifically I'm working with (12-10x)^-1. Using normal antideriv rules it would just be (12-10x)^0/0, which is irrational, could someone answer this for me? Thanks
    antiderivative involves the ln function.

     <br />
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C<br />

    normally, one learns how to find the derivative of ln functions before tackling antiderivatives.
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    Quote Originally Posted by skeeter View Post
    antiderivative involves the ln function.

     <br />
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C<br />

    normally, one learns how to find the derivative of ln functions before tackling antiderivatives.
    Yeah well my calc I class was a joke, I passed out with an A and now I'm struggling hard with calc II. thanks for the reply though

    Quote Originally Posted by skeeter View Post
    antiderivative involves the ln function.

     <br />
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C<br />

    normally, one learns how to find the derivative of ln functions before tackling antiderivatives.
    Where did the -1/10 come from in your solution?
    Last edited by Krizalid; March 10th 2009 at 12:06 PM.
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    Quote Originally Posted by fattydq View Post
    Yeah well my calc I class was a joke, I passed out with an A and now I'm struggling hard with calc II. thanks for the reply though



    Where did the -1/10 come from in your solution?
    He made the substitution u= 12- 10x so du= -10 dx so dx= -(1/10)du.
    \int \frac{dx}{12- 10x}= \int\frac{-(1/10)du}{u}= -\frac{1}{10}\int\frac{du}{u}

    = -\frac{1}{10}ln|u|+ C= -\frac{1}{10}ln|12-10x|+ C
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