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(D) Given the differential equation: $(1+x^{2}) dt = (	an^{-1} x - t) dx$. Divide both sides by $(1+x^{2}) dx$: $\frac{dt}{dx} = \frac{	an^{-1} x -

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$e^{	an^{-1} x}$

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(D) Given the differential equation: $(1+x^{2}) dt = (	an^{-1} x - t) dx$. Divide both sides by $(1+x^{2}) dx$: $\frac{dt}{dx} = \frac{	an^{-1} x - t}{1+x^{2}}$ Rearrange the equation into the standard linear form $\frac{dt}{dx} + P(x)t = Q(x)$: $\frac{dt}{dx} + \frac{1}{1+x^{2}}t = \frac{	an^{-1} x}{1+x^{2}}$ Here,$P(x) = \frac{1}{1+x^{2}}$. The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx}$: $I.F. = e^{\int \frac{1}{1+x^{2}} dx} = e^{	an^{-1} x}$.

Find the integrating factor of the differential equation $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.

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