For x 0,let f(x) = int_{1}^{x} frac{log t}{ | vedclass

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(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt$.Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt$.Let $t = \frac{1}{u}$,

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$\frac{1}{2}(\log x)^2$

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(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt$.Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt$.Let $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$. When $t=1, u=1$ and when $t=1/x, u=x$.$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln(1/u)}{1 + 1/u} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{\frac{u+1}{u}} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{\ln u}{u(u+1)} du$.Now,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{x} \frac{\ln t}{t(1+t)} dt = \int_{1}^{x} \ln t \left( \frac{1}{1+t} + \frac{1}{t(1+t)} \right) dt$.Simplifying the integrand: $\frac{1}{1+t} + \frac{1}{t(1+t)} = \frac{t+1}{t(1+t)} = \frac{1}{t}$.Thus,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{t} dt$.Let $\ln t = v$,then $\frac{1}{t} dt = dv$. When $t=1, v=0$ and when $t=x, v=\ln x$.$f(x) + f\left(\frac{1}{x}\right) = \int_{0}^{\ln x} v dv = \left[ \frac{v^2}{2} \right]_{0}^{\ln x} = \frac{1}{2}(\ln x)^2$.

For $x > 0$,let $f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt$. Then $f(x) + f\left(\frac{1}{x}\right)$ is equal to:

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