I 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.
I still do not understand how to solve the following problem. Please show step-by-step.
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!!!!!!!
2 answers
"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.
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.
4 answers