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integrate : e2x - 1 / e2x + 1

  • 34
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If you know the answer you should simply differentiate the answer and see if it fits.

Anyway,
∫(exp(2x) - 1)/(exp(2x) + 1) dx =
∫(exp(x) - exp(-x))/(exp(x) + exp(-x)) dx =
∫(exp(x) - exp(-x))/2 /(exp(x) + exp(-x))/2 dx =
∫sh(x) / ch(x) dx =
∫th(x) dx =
ln(ch(x))

  • 5

The given equation is (e^2x - 1) / (e^2x + 1).
Now divide the given equation by e^x.
the equation reduces to (e^x - e^-x)/(e^x + e^-x)

put e^x + e^-x = p
then differentiating on both sides gives (e^x - e^-x )dx = dp

thus the given equation can be re written as dp/p
integrating on both sides gives ln(p) + k or log(p) + k
where k is any arbitrary constant or also known as constant of integration.

thus the final answer is log(e^x + e^-x ) + k

  • 5
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