If f(x) = 1 + x + int_{1}^{x} (ln^2 t + 2 ln | vedclass

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(A) The domain of $f(x)$ is $x > 0$ because the logarithmic function $\ln t$ is defined only for $t > 0$.Given $f(x) = 1 + x + \int_{1}^{x} (\ln^2 t +

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$(0, \infty)$

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(A) The domain of $f(x)$ is $x > 0$ because the logarithmic function $\ln t$ is defined only for $t > 0$.Given $f(x) = 1 + x + \int_{1}^{x} (\ln^2 t + 2 \ln t) \, dt$.Differentiating both sides with respect to $x$ using the Leibniz integral rule:$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx} \int_{1}^{x} (\ln^2 t + 2 \ln t) \, dt$$f'(x) = 0 + 1 + (\ln^2 x + 2 \ln x)$$f'(x) = \ln^2 x + 2 \ln x + 1$$f'(x) = (\ln x + 1)^2$Since $(\ln x + 1)^2 \ge 0$ for all $x > 0$,and $f'(x) = 0$ only at $x = e^{-1}$,the function $f(x)$ is strictly increasing on its entire domain $(0, \infty)$.

If $f(x) = 1 + x + \int_{1}^{x} (\ln^2 t + 2 \ln t) \, dt$,then $f(x)$ increases in

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