It is given that frac{d}{dt}(t log t - t) = l | vedclass

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(C) Given that $\frac{d}{dt}(t \log t - t) = \log t$. Let $I = \int_0^1 2x \log(1+x^2) dx$. Substitute $t = 1+x^2$, then $dt = 2x dx$. When $x=0$, $t=

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$\frac{4}{e}$

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(C) Given that $\frac{d}{dt}(t \log t - t) = \log t$. Let $I = \int_0^1 2x \log(1+x^2) dx$. Substitute $t = 1+x^2$, then $dt = 2x dx$. When $x=0$, $t=1$. When $x=1$, $t=2$. Thus, $I = \int_1^2 \log t dt$. Using the given derivative, $\int \log t dt = t \log t - t + C$. So, $I = [t \log t - t]_1^2 = (2 \log 2 - 2) - (1 \log 1 - 1) = 2 \log 2 - 2 - 0 + 1 = \log 4 - 1$. Therefore, $\exp(I) = \exp(\log 4 - 1) = e^{\log 4} \cdot e^{-1} = 4 \cdot \frac{1}{e} = \frac{4}{e}$. Thus, the correct option is $C$.

It is given that $\frac{d}{dt}(t \log t - t) = \log t$. Then, $\exp \left( \int_0^1 2x \log(1+x^2) dx \right) = $

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