int {frac{{{{cot }^{ - 1}}({e^x})}}{{{e^x}}}} | vedclass

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Question

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

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 $e^x = t$,then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.Therefore,$I = \int \frac{\cot ^{-1}(t)}{t^2} dt$.Using integration by parts,let $u = \cot ^{-1}(t)$ and $dv = t^{-2} dt$.Then $du = -\frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$.$I = u v - \int v du = -\frac{\cot ^{-1}(t)}{t} - \int \left(-\frac{1}{t}\right) \left(-\frac{1}{1+t^2}\right) dt$.$I = -\frac{\cot ^{-1}(t)}{t} - \int \frac{1}{t(1+t^2)} dt$.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,$I = -\frac{\cot ^{-1}(t)}{t} - (\ln|t| - \frac{1}{2} \ln(1+t^2)) + C$.$I = \frac{1}{2} \ln(1+t^2) - \ln|t| - \frac{\cot ^{-1}(t)}{t} + C$.Since $t = e^x$,$\ln|t| = x$.$I = \frac{1}{2} \ln(1+e^{2x}) - x - \frac{\cot ^{-1}(e^x)}{e^x} + C$.

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

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