of int_(0)^(ln 3)(e^(-x))/(1+e^(-x))dx

Substitution and Simplification: Simplify the integrand by using a substitution. Let u = 1 + e^(-x), which implies that du = -e^(-x)dx. We need to express dx in terms of du.Calculate dx: Calculate dx from the substitution. du = -e^(-x)dx implies dx = -du/e^(-x). Now we can substitute this into the integral.Change Limits of Integration: Change the limits of integration according to the substitution. When x = 0, u = 1 + e^0 = 2. When x = ln(3), u = 1 + e^(-ln(3)) = 1 + 1/3 = 4/3.Rewrite Integral with New Variable: Rewrite the integral with the new variable and limits. The integral becomes int_(2)^(4/3)(-du/u). The negative sign comes from the dx substitution.Evaluate Integral: Evaluate the integral with the new variable. The integral of -1/u du is -ln|u|. So we have -ln|u| evaluated from 2 to 4/3.Apply Fundamental Theorem of Calculus: Apply the Fundamental Theorem of Calculus. We need to calculate -ln|u| from 2 to 4/3, which is -ln(4/3) + ln(2).Simplify Result: Simplify the result. -ln(4/3) + ln(2) = ln(2) - ln(4/3) = ln(2) - ln(4) + ln(3) = ln(2/4) + ln(3) = ln(1/2) + ln(3) = ln(3/2).Check for Errors: Check for any mathematical errors in the previous steps. No mathematical errors were made in the previous steps.

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Evaluate the limit: $\int_{0}^{\ln 3} \frac{\mathrm{e}^{-x}}{1+\mathrm{e}^{-x}} \mathrm{~d} x$

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Evaluate definite integrals using the chain rule

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Q. Evaluate the limit:

\int_{0}^{\ln 3} \frac{\mathrm{e}^{-x}}{1+\mathrm{e}^{-x}} \mathrm{~d} x

Substitution and Simplification: Simplify the integrand by using a substitution. Let $u = 1 + e^{-x}$ , which implies that $du = -e^{-x}dx$ . We need to express $dx$ in terms of $du$ .
Calculate $dx$ : Calculate $dx$ from the substitution. $du = -e^{-x}dx$ implies $dx = -\frac{du}{e^{-x}}$ . Now we can substitute this into the integral.
Change Limits of Integration: Change the limits of integration according to the substitution. $\newline$ When $x = 0$ , $u = 1 + e^0 = 2$ . $\newline$ When $x = \ln(3)$ , $u = 1 + e^{-\ln(3)} = 1 + \frac{1}{3} = \frac{4}{3}$ .
Rewrite Integral with New Variable: Rewrite the integral with the new variable and limits. $\newline$ The integral becomes $\int_{2}^{\frac{4}{3}}\left(-\frac{du}{u}\right)$ . The negative sign comes from the $dx$ substitution.
Evaluate Integral: Evaluate the integral with the new variable. $\newline$ The integral of $-\frac{1}{u} \, du$ is $-\ln|u|$ . So we have $-\ln|u|$ evaluated from $2$ to $\frac{4}{3}$ .
Apply Fundamental Theorem of Calculus: Apply the Fundamental Theorem of Calculus. We need to calculate $-\ln|u|$ from $2$ to $\frac{4}{3}$ , which is $-\ln(\frac{4}{3}) + \ln(2)$ .
Simplify Result: Simplify the result. $-\ln(\frac{4}{3}) + \ln(2) = \ln(2) - \ln(\frac{4}{3}) = \ln(2) - \ln(4) + \ln(3) = \ln(\frac{2}{4}) + \ln(3) = \ln(\frac{1}{2}) + \ln(3) = \ln(\frac{3}{2})$ .
Check for Errors: Check for any mathematical errors in the previous steps. $\newline$ No mathematical errors were made in the previous steps.