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| Math Central | Quandaries & Queries |
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Question from Indrajit, a student: ∫4e^x + 6e^-x/(9e^x + 4e^-x)dx = Ax + Bloge(9e2x - 4) + C then A=?......B=?.....C=? plz solve it...."^" stands for "to the power of".... |
Indrajit,
I think the integrand should be (4ex + 6e-x)/(9ex + 4e-x). Also check the right side, should it be Ax + Bloge(9e2x + 4) + C?
My approach to this problem would be to use the Fundamental Theorem of Calculus. Since the right side is an antiderivative of the integrand, the derivative of the right side must be the integrand. Differentiate the right side and set the derivative equal to the integrand. These must be identical functions which should allow you to solve for A and B.
I hope this helps,
Harley
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