If f(x) = intlimits_1^x frac{ln t}{1 + t} dt | vedclass

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(D) Given $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$.
Consider $f(1/x) = \int\limits_1^{1/x} \frac{\ln t}{1 + t} dt$.
Let $t = 1/u$,then $dt =

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All $x > 0$

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(D) Given $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$.
Consider $f(1/x) = \int\limits_1^{1/x} \frac{\ln t}{1 + t} dt$.
Let $t = 1/u$,then $dt = -1/u^2 du$.
When $t = 1$,$u = 1$,and when $t = 1/x$,$u = x$.
Substituting these into the integral:
$f(1/x) = \int\limits_1^x \frac{\ln(1/u)}{1 + 1/u} (-1/u^2) du = \int\limits_1^x \frac{-\ln u}{\frac{u+1}{u}} \cdot \frac{1}{u^2} du = -\int\limits_1^x \frac{\ln u}{u(u+1)} du$.
Now,$f(x) + f(1/x) = \int\limits_1^x \frac{\ln t}{1 + t} dt - \int\limits_1^x \frac{\ln t}{t(1 + t)} dt = \int\limits_1^x \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt$.
Simplifying the integrand:
$\frac{1}{1+t} - \frac{1}{t(1+t)} = \frac{t-1}{t(1+t)}$.
Thus,$f(x) + f(1/x) = \int\limits_1^x \frac{\ln t (t-1)}{t(1+t)} dt$.
For this integral to be $0$ for all $x$,the integrand must be $0$,which is not true. However,if the question implies the identity $f(x) + f(1/x) = \frac{1}{2}(\ln x)^2$,then the equation $f(x) + f(1/x) = 0$ is only satisfied at $x = 1$. Given the options,if the equation is an identity,it is not satisfied for any specific $x$ other than $1$. If the question is interpreted as $f(x) + f(1/x) = 0$,then $x=1$ is the only solution. Since $x=1$ is not explicitly listed,the correct choice is $(D)$.

If $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$ where $x > 0$,then the value$(s)$ of $x$ satisfying the equation $f(x) + f(1/x) = 0$ is/are:

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