If f(t) = int_{-t}^{t} frac{dx}{1 + x^2},then | vedclass

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(D) Given $f(t) = \int_{-t}^{t} \frac{dx}{1 + x^2}$.We know that $\int \frac{dx}{1 + x^2} = 	an^{-1}(x)$.Applying the limits:$f(t) = [	an^{-1}(x)]_{

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(D) Given $f(t) = \int_{-t}^{t} \frac{dx}{1 + x^2}$.We know that $\int \frac{dx}{1 + x^2} = 	an^{-1}(x)$.Applying the limits:$f(t) = [	an^{-1}(x)]_{-t}^{t} = 	an^{-1}(t) - 	an^{-1}(-t)$.Since $	an^{-1}(-t) = -	an^{-1}(t)$,we have:$f(t) = 	an^{-1}(t) - (-	an^{-1}(t)) = 2	an^{-1}(t)$.Differentiating $f(t)$ with respect to $t$ using the chain rule:$f'(t) = 2 	imes \frac{1}{1 + t^2} = \frac{2}{1 + t^2}$.Now,substitute $t = 1$:$f'(1) = \frac{2}{1 + (1)^2} = \frac{2}{2} = 1$.

If $f(t) = \int_{-t}^{t} \frac{dx}{1 + x^2}$,then $f'(1)$ is

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