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

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(A) To evaluate the integral $I = \int \frac{1}{e^x+1} dx$,we can rewrite the integrand as follows:$I = \int \frac{1}{e^x(1 + e^{-x})} dx = \int \frac

Vedclass · Accepted Answer

$\log \left|\frac{e^x}{e^x+1}\right|$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{1}{e^x+1} dx$,we can rewrite the integrand as follows:$I = \int \frac{1}{e^x(1 + e^{-x})} dx = \int \frac{e^{-x}}{1 + e^{-x}} dx$.Let $u = 1 + e^{-x}$. Then $du = -e^{-x} dx$,which implies $e^{-x} dx = -du$.Substituting these into the integral,we get:$I = \int \frac{-du}{u} = -\log |u| + C = -\log |1 + e^{-x}| + C$.We can simplify this as:$I = -\log \left| \frac{e^x + 1}{e^x} \right| + C = -(\log |e^x + 1| - \log |e^x|) + C = -\log |e^x + 1| + x + C$.Alternatively,using the form $I = \int \frac{1}{e^x+1} dx = \int \frac{e^x+1-e^x}{e^x+1} dx = \int (1 - \frac{e^x}{e^x+1}) dx = x - \log |e^x+1| + C$.Since $x = \log(e^x)$,we have $I = \log(e^x) - \log |e^x+1| + C = \log \left| \frac{e^x}{e^x+1} \right| + C$.Thus,the correct option is $A$.

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

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