int frac{cot^{-1}(e^x)}{e^x} dx is equal to: | vedclass

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(C) Let $I = \int \frac{\cot^{-1}(e^x)}{e^x} dx$.Substitute $e^x = t$,so $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.Then $I = \int \frac{\cot^{-1

Vedclass · Accepted Answer

$\frac{1}{2} \ln(e^{2x} + 1) - \frac{\cot^{-1}(e^x)}{e^x} - x + c$

Vedclass · Answer

(C) Let $I = \int \frac{\cot^{-1}(e^x)}{e^x} dx$.Substitute $e^x = t$,so $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.Then $I = \int \frac{\cot^{-1}(t)}{t^2} dt$.Using integration by parts,let $u = \cot^{-1}(t)$ and $dv = t^{-2} dt$. Then $du = -\frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$.$I = -\frac{\cot^{-1}(t)}{t} - \int \left(-\frac{1}{t}\right) \left(-\frac{1}{1+t^2}\right) dt = -\frac{\cot^{-1}(t)}{t} - \int \frac{1}{t(1+t^2)} dt$.Using partial fractions: $\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.So,$\int \frac{1}{t(1+t^2)} dt = \int \frac{1}{t} dt - \int \frac{t}{1+t^2} dt = \ln|t| - \frac{1}{2} \ln(1+t^2)$.Substituting back into $I$: $I = -\frac{\cot^{-1}(t)}{t} - (\ln|t| - \frac{1}{2} \ln(1+t^2)) + C = \frac{1}{2} \ln(1+t^2) - \ln|t| - \frac{\cot^{-1}(t)}{t} + C$.Since $t = e^x$,$\ln|t| = x$ and $t^2 = e^{2x}$.$I = \frac{1}{2} \ln(1+e^{2x}) - x - \frac{\cot^{-1}(e^x)}{e^x} + C$.

$\int \frac{\cot^{-1}(e^x)}{e^x} dx$ is equal to:

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