If int {frac{{{x^2} - x + 1}}{{{x^2} + 1}}{e^ | vedclass

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(B) Let $I = \int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} \cdot {e^{{{\cot }^{ - 1}}x}}dx$.Substitute $x = \cot t$,then $dx = -\csc^2 t \, dt$.Since $1

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$x$

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(B) Let $I = \int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} \cdot {e^{{{\cot }^{ - 1}}x}}dx$.Substitute $x = \cot t$,then $dx = -\csc^2 t \, dt$.Since $1 + \cot^2 t = \csc^2 t$,the integral becomes:$I = \int {\frac{{\cot^2 t - \cot t + 1}}{{\csc^2 t}}} \cdot {e^t} \cdot (-\csc^2 t) \, dt$$I = - \int {e^t} (\cot^2 t - \cot t + 1) \, dt$$I = - \int {e^t} (\csc^2 t - \cot t) \, dt$$I = \int {e^t} (\cot t - \csc^2 t) \, dt$Using the formula $\int {e^t} (f(t) + f'(t)) \, dt = {e^t} f(t) + C$,where $f(t) = \cot t$ and $f'(t) = -\csc^2 t$:$I = {e^t} \cot t + C$Substituting $t = \cot^{-1} x$ back:$I = {e^{\cot^{-1} x}} \cdot x + C$Comparing this with $A(x) {e^{\cot^{-1} x}} + C$,we get $A(x) = x$.

If $\int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}{e^{{{\cot }^{ - 1}}x}}dx = A(x) {e^{{{\cot }^{ - 1}}x}} + C}$,then $A(x)$ is equal to

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