The value of the definite integral int_{1}^{i | vedclass

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(A) Let $I = \int_{1}^{\infty} \frac{dx}{e \cdot e^x + e^3 \cdot e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_{1}^{\infty} \frac

Vedclass · Accepted Answer

$\frac{\pi}{4e^2}$

Vedclass · Answer

(A) Let $I = \int_{1}^{\infty} \frac{dx}{e \cdot e^x + e^3 \cdot e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_{1}^{\infty} \frac{e^x \, dx}{e(e^{2x} + e^2)}$.Substitute $e^x = t$,so $e^x \, dx = dt$. When $x=1, t=e$; when $x 	o \infty, t 	o \infty$.$I = \frac{1}{e} \int_{e}^{\infty} \frac{dt}{t^2 + e^2}$.Using the formula $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} 	an^{-1} \frac{t}{a}$:$I = \frac{1}{e} \left[ \frac{1}{e} 	an^{-1} \frac{t}{e} ight]_{e}^{\infty} = \frac{1}{e^2} \left[ 	an^{-1}(\infty) - 	an^{-1}(1) ight]$.$I = \frac{1}{e^2} \left( \frac{\pi}{2} - \frac{\pi}{4} ight) = \frac{\pi}{4e^2}$.

The value of the definite integral $\int_{1}^{\infty} (e^{x+1} + e^{3-x})^{-1} \, dx$ is

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