六、解答题 (本题满分 9 分)
设 $f(x)=\int_{1}^{x} \frac{\ln t}{1+t} \mathrm{~d} t$, 其中 $x>0$, 求 $f(x)+f\left(\frac{1}{x}\right)$.
注意：解答此类问题，先别想着有什么技巧，先按照题目所问，算几步看一看。
$$
F(x)=f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln t}{1+t} \mathrm{d} t+\int_{1}^{\frac{1}{x}} \frac{\ln t}{1+t} \mathrm{d} x
$$
变限积分问题一般都是要求导：
$$
F^{\prime}(x)=\frac{\ln x}{1+x}-\frac{1}{x^{2}} \frac{\ln \frac{1}{x}}{1+\frac{1}{x}} \Rightarrow
$$
$$
F^{\prime}(x)=\frac{x \ln x}{x(1+x)}+\frac{\ln x}{x(1+x)}=\frac{\ln x(x+1)}{x(1+x)}
$$
$$
F^{\prime}(x)=\frac{\ln x}{x} \Rightarrow
$$
$$
F(x)=f(x)+f\left(\frac{1}{x}\right)=\int F^{\prime}(x) \mathrm{d} x =
$$
$$
\int \frac{\ln x}{x} \mathrm{d} x=\int \ln x \mathrm{d} (\ln x)=\frac{1}{2} \ln ^{2} x+C
$$
或者：
$$
f\left(\frac{1}{x}\right)=\int_{1}^{\frac{1}{x}} \frac{\ln t}{1+t} \mathrm{d} t \Rightarrow t=\frac{1}{u} \Rightarrow
$$
$$
t \in\left(1, \frac{1}{x}\right) \Rightarrow u \in(1, x) \Rightarrow \mathrm{d} t=\frac{-1}{u^{2}} \mathrm{~ d} u \Rightarrow
$$
$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln \frac{1}{u}}{1+\frac{1}{u}} \cdot \frac{-1}{u^{2}} \mathrm{~ d} u=
$$
$$
\int_{1}^{x} \frac{0-\ln u}{1+u} \cdot \frac{-1}{u} \mathrm{~ d} u=\int_{1}^{x} \frac{\ln u}{u(1+u)} \mathrm{~ d} u=
$$
$$
\int_{1}^{x} \frac{\ln t}{t(1+t)} \mathrm{d} t \Rightarrow
$$
于是：
$$
F(x)=f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln t}{1+t} \mathrm{d} t t \int_{1}^{x} \frac{\ln t}{t(1+t)} \mathrm{d} t=
$$
$$
\int_{1}^{x} \frac{\ln t(t+1)}{t(1+t)} \mathrm{d} t=\int_{1}^{x} \frac{\ln t}{t}=\frac{1}{2} \ln ^{2} x+C
$$