问题及解答

设 $f(x)=\int_1^x\frac{\ln t}{1+t}dt$, 求 $f(x)+f(\frac{1}{x})$.

令 $g(x)=f(x)+f(\frac{1}{x})$, 则 $g(1)=2f(1)=0$, 且

\[
g'(x)=\frac{\ln x}{1+x}+\frac{\ln\frac{1}{x}}{1+\frac{1}{x}}\cdot\frac{-1}{x^2}=\frac{\ln x}{x}.
\]

于是就是求微分方程

\[
\begin{cases}
g'(x)=\frac{\ln x}{x},\\
g(1)=0.
\end{cases}
\]

因此,

\[
g(x)=\int_1^x\frac{\ln t}{t}dt=\int_1^x\ln t d\ln t=(\ln t)^2\biggr|_{1}^{x}-\int_1^x\ln t d\ln t,
\]

这推出

\[
2g(x)=(\ln x)^2,
\]

即 $g(x)=\frac{1}{2}(\ln x)^2$.

（法二）

\[
\begin{split}
f(\frac{1}{x})&=\int_{1}^{\frac{1}{x}}\frac{\ln t}{1+t}\mathrm{d}t\\
&=\int_{1}^{x}\frac{\ln\frac{1}{u}}{1+\frac{1}{u}}\cdot\frac{-1}{u^2}\mathrm{d}u\\
&=\int_{1}^{x}\frac{\ln u}{u^2+u}\mathrm{d}u
\end{split}
\]

于是

\[
\begin{split}
f(x)+f(\frac{1}{x})&=\int_{1}^{x}\frac{\ln t}{1+t}\mathrm{d}t+\int_{1}^{x}\frac{\ln t}{t^2+t}\mathrm{d}t\\
&=\int_{1}^{x}\frac{\ln t+t\ln t}{t(1+t)}\mathrm{d}t\\
&=\int_{1}^{x}\frac{\ln t}{t}\mathrm{d}t\\
&=\int_{1}^{x}\ln t\mathrm{d}(\ln t)\\
&=\frac{1}{2}(\ln t)^2\biggr|_{1}^{x}\\
&=\frac{1}{2}(\ln x)^2
\end{split}
\]