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.∫(e 200x +e 202x)/ (e x +e -x )dx.

 

.∫(e200x+e202x)/(ex+e-x)dx.

 

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Sudent,
Please find answer to your question below

I = \int \frac{e^{200x}+e^{202x}}{e^{x}+e^{-x}}dx
I = \int \frac{e^{200x}}{e^{-x}}.\frac{1+e^{2x}}{e^{2x}+1}dx
I = \int e^{201x}.dx
I = \frac{e^{201x}}{201} + constant

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