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

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(B) Let $I = \int \frac{dx}{e^x+e^{-x}+2}$.We can rewrite the denominator as:$e^x + \frac{1}{e^x} + 2 = \frac{e^{2x} + 1 + 2e^x}{e^x} = \frac{(e^x+1)^

Vedclass · Accepted Answer

$\frac{-1}{e^x+1}+c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{e^x+e^{-x}+2}$.We can rewrite the denominator as:$e^x + \frac{1}{e^x} + 2 = \frac{e^{2x} + 1 + 2e^x}{e^x} = \frac{(e^x+1)^2}{e^x}$.Therefore,the integral becomes:$I = \int \frac{e^x}{(e^x+1)^2} dx$.Let $u = e^x + 1$. Then $du = e^x dx$.Substituting these into the integral:$I = \int \frac{du}{u^2} = \int u^{-2} du$.Integrating with respect to $u$:$I = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c$.Substituting back $u = e^x + 1$:$I = -\frac{1}{e^x+1} + c$.

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

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