The function F(x) = int_0^x log left( frac{1 | vedclass

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(A) Let $f(t) = \log \left( \frac{1 - t}{1 + t} \right)$.First,we check if $f(t)$ is an odd function by evaluating $f(-t)$:$f(-t) = \log \left( \frac{

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An even function

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(A) Let $f(t) = \log \left( \frac{1 - t}{1 + t} \right)$.First,we check if $f(t)$ is an odd function by evaluating $f(-t)$:$f(-t) = \log \left( \frac{1 - (-t)}{1 + (-t)} \right) = \log \left( \frac{1 + t}{1 - t} \right) = \log \left( \left( \frac{1 - t}{1 + t} \right)^{-1} \right) = -\log \left( \frac{1 - t}{1 + t} \right) = -f(t)$.Since $f(-t) = -f(t)$,the function $f(t)$ is an odd function.We know that if $f(t)$ is an odd function,then the integral $F(x) = \int_0^x f(t) \,dt$ is an even function.To verify,$F(-x) = \int_0^{-x} f(t) \,dt$. Let $t = -u$,then $dt = -du$. When $t=0, u=0$; when $t=-x, u=x$.$F(-x) = \int_0^x f(-u) (-du) = \int_0^x (-f(u)) (-du) = \int_0^x f(u) \,du = F(x)$.Since $F(-x) = F(x)$,$F(x)$ is an even function.

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

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