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,

Question

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

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

Answer 1

Given,

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

Aliter :