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September 16, 2014

September 16, 2014

Posted by **Anthony** on Saturday, December 8, 2007 at 11:49pm.

x = Integral (0 to v) dv/(q^2-v^2) where q = constant

the answer is --> x = 1/2q ln (q+v/q-v)

1) Where 1/2q came from?

2) Where ln came from? Is there a formula for it?

Please help!!!!!!!

- Math -
**drwls**, Sunday, December 9, 2007 at 12:46amI explained this yesterday. You use the method of partial fractions, to rewrite

1/(q^2-v^2) as the sum of terms with

(q+v)and (q-v) in the denominator. When you integrate those two separate terms, you get the difference of two log terms.

Look up the method in your text or verify the integal in a table of integrals.

[1/(2q)] is a factor appears when the method of partial fractions is correctly applied.

- Math -
**drwls**, Sunday, December 9, 2007 at 4:51am"ln" is term that designates the natural (base e) logarithm of whatever follows. ln x is the integral of dx/x

ln (q+v) is the integral of dv/(q+v), with q being a constant. When you do the integration, you get the ln(q+v) - ln (q-v). Using the rules of logarithms, this can be written ln[(q+v)/(q-v)]. The 1/2q factor comes from the step of converting 1/(q^2-v^2) to two partial fractions.

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