If ∫(2e^x+3e^(-x))/(4e^x+7e^(-x))dx = 1/14 ux + v loge(4e^x + 7e^(–x))+ C where C is a constant of integration,

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Asked: 2023-03-06T20:55:26+05:30 In: Other
If ∫(2e^x+3e^(-x))/(4e^x+7e^(-x))dx = 1/14 ux + v loge(4e^x + 7e^(–x))+ C where C is a constant of integration,

Hi, I have a question and I hope anyone could answer it:

If (intfrac{2e^x+3e^-x}{4e^x+7e^{-x}}dx) = (frac{1}{14})(ux + v loge(4ex + 7e–x))+ C, where C is a constant of integration, then u + v is equal to __________.

∫(2ex+3e-x)/(4ex+7e-x)dx = 1/14 ux + v loge(4ex + 7e–x))+ C

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1 Answer

  1. Best Answer
    Gandalf

    Answer:

    (2ex+3e-x)/(4ex+7e-x)dx = 1/14 ux + v loge(4ex + 7e–x))+ C 

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