If int e^{-x} tan ^{-1}left(e^xright) d x = f | vedclass

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(D) Let $I = \int e^{-x} 	an ^{-1}\left(e^xight) d x$. Substitute $e^x = t$,then $e^x d x = d t$,which implies $d x = \frac{1}{t} d t$. Substitutin

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$x - e^{-x} 	an ^{-1}\left(e^xight)$

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(D) Let $I = \int e^{-x} 	an ^{-1}\left(e^xight) d x$. Substitute $e^x = t$,then $e^x d x = d t$,which implies $d x = \frac{1}{t} d t$. Substituting these into the integral,we get: $I = \int \frac{	an ^{-1} t}{t^2} d t$. Using integration by parts,let $u = 	an ^{-1} t$ and $dv = t^{-2} dt$. Then $du = \frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$. $I = -\frac{1}{t} 	an ^{-1} t - \int \left(-\frac{1}{t}ight) \frac{1}{1+t^2} d t = -\frac{1}{t} 	an ^{-1} t + \int \frac{1}{t(1+t^2)} d t$. Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$. So,$I = -\frac{1}{t} 	an ^{-1} t + \int \left(\frac{1}{t} - \frac{t}{1+t^2}ight) d t$. $I = -\frac{1}{t} 	an ^{-1} t + \log |t| - \frac{1}{2} \log (1+t^2) + C$. Substituting $t = e^x$ back: $I = -e^{-x} 	an ^{-1} (e^x) + \log (e^x) - \frac{1}{2} \log (1+e^{2x}) + C$. Since $\log (e^x) = x$,we have $I = -e^{-x} 	an ^{-1} (e^x) + x - \frac{1}{2} \log (1+e^{2x}) + C$. Comparing this with the given expression $f(x) - \frac{1}{2} \log (1+e^{2x}) + C$,we find $f(x) = x - e^{-x} 	an ^{-1} (e^x)$.

If $\int e^{-x} \tan ^{-1}\left(e^x\right) d x = f(x) - \frac{1}{2} \log \left(1+e^{2 x}\right) + C$,then $f(x)$ equals

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