For x > 0, if \(f(x)=\displaystyle\int_1^x \frac{\log_e t}{(1+t)}\,dt\), then f(e) + f(1/e) equals
$\frac{1}{2}$
In \(f(1/e)=\displaystyle\int_1^{1/e}\dfrac{\log_e t}{1+t}\,dt\), substitute \(t=1/u\), so \(dt=-du/u^2\); the limits \(t=1\to 1/e\) become \(u=1\to e\).
Since \(\log_e t=-\log_e u\) and \(1+t=\dfrac{u+1}{u}\), the integrand becomes \(\dfrac{-\log_e u\cdot u}{u+1}\); multiplying by \(dt=-du/u^2\) gives \(f(1/e)=\displaystyle\int_1^e \dfrac{\log_e u}{u(u+1)}\,du\).
Adding \(f(e)=\displaystyle\int_1^e \dfrac{\log_e u}{1+u}\,du\) to this: \(f(e)+f(1/e)=\displaystyle\int_1^e \log_e u\left[\dfrac{1}{1+u}+\dfrac{1}{u(1+u)}\right]du\).
The bracket simplifies: \(\dfrac{1}{1+u}+\dfrac{1}{u(1+u)}=\dfrac{u+1}{u(1+u)}=\dfrac{1}{u}\), so \(f(e)+f(1/e)=\displaystyle\int_1^e \dfrac{\log_e u}{u}\,du\).
This is a standard integral: \(\displaystyle\int_1^e \dfrac{\log_e u}{u}\,du=\left[\dfrac{(\log_e u)^2}{2}\right]_1^e=\dfrac{1^2}{2}-0=\dfrac12\).
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