If int frac{dx}{e^x + 4e^{-x}} = f(x) + c,the | vedclass

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(A) Given integral is $I = \int \frac{dx}{e^x + 4e^{-x}}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x dx}{e^{2x} + 4}$.Let $u = e^

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$\frac{1}{2} 	an^{-1}(\frac{e^x}{2})$

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(A) Given integral is $I = \int \frac{dx}{e^x + 4e^{-x}}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x dx}{e^{2x} + 4}$.Let $u = e^x$,then $du = e^x dx$.The integral becomes $I = \int \frac{du}{u^2 + 2^2}$.Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{2} 	an^{-1}(\frac{u}{2}) + c$.Substituting $u = e^x$ back,we get $I = \frac{1}{2} 	an^{-1}(\frac{e^x}{2}) + c$.Thus,$f(x) = \frac{1}{2} 	an^{-1}(\frac{e^x}{2})$.

If $\int \frac{dx}{e^x + 4e^{-x}} = f(x) + c$,then $f(x)$ is

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