int frac{e^{2x}-1}{e^{2x}+1} dx = . . . . . | vedclass

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Question

(A) To solve the integral $I = \int \frac{e^{2x}-1}{e^{2x}+1} dx$,we can rewrite the integrand by dividing the numerator and denominator by $e^x$:$I =

Vedclass · Accepted Answer

$\log(e^{2x}+1) - x$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{e^{2x}-1}{e^{2x}+1} dx$,we can rewrite the integrand by dividing the numerator and denominator by $e^x$:$I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$Let $u = e^x + e^{-x}$. Then,the derivative is $du = (e^x - e^{-x}) dx$.Substituting these into the integral,we get:$I = \int \frac{1}{u} du = \log|u| + C$Substituting $u$ back,we have $I = \log|e^x + e^{-x}| + C$.Alternatively,we can write $e^x + e^{-x} = \frac{e^{2x}+1}{e^x}$.So,$I = \log|\frac{e^{2x}+1}{e^x}| + C = \log|e^{2x}+1| - \log|e^x| + C = \log(e^{2x}+1) - x + C$.Comparing this with the given options,the correct choice is $A$.

$\int \frac{e^{2x}-1}{e^{2x}+1} dx = $ . . . . . . $+ C$.

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