对变限积分做求导运算之后，要再通过积分运算变回来的话，需要保持原本的积分上下限不变

一、题目

若函数 $f(x)$ $=$ $\int_{1}^{x} \frac{\ln t}{1+t} \mathrm{~d} t$, $x > 0$, 则 $f(x)+f\left(\frac{1}{x}\right)$ $=$ $?$

难度评级：

二、解析

方法 1：先求导再积分

$$
F(x)=f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln t}{1+t} \mathrm{~d} t+\int_{1}^{\frac{1}{x}} \frac{\ln t}{1+t} \mathrm{~d} t
$$

$$
F^{\prime}(x)=f^{\prime}(x)+f^{\prime}\left(\frac{1}{x}\right) \cdot\left(\frac{-1}{x^{2}}\right) \Rightarrow
$$

注意：这里在对变限积分求导的时候可以直接进行，不需要对 $1+x$ 做变量替换，原因可以参考下面两篇文章：

1. 有关变限积分求导的几种形式
2. 什么情况下可以把 x 看作常数

$$
F^{\prime}(x)=\frac{\ln x}{1+x}+\frac{\ln \frac{1}{x}}{1+\frac{1}{x}}\left(\frac{-1}{x^{2}}\right) \Rightarrow
$$

$$
F^{\prime}(x)=\frac{\ln x}{1+x}+\frac{\ln x^{-1}}{-x^{2}\left(\frac{x+1}{x}\right)} \Rightarrow
$$

$$
F^{\prime}(x)=\frac{\ln x}{1+x}+\frac{\ln x}{x(1+x)} \Rightarrow
$$

$$
F^{\prime}(x)=\frac{x \ln x + \ln x}{x(1+x)}
$$

$$
F^{\prime}(x)=\frac{\ln x}{x} \Rightarrow
$$

$$
F(x)=\int_{1}^{x} F^{\prime}(t) \mathrm{~d} t=\int_{1}^{x} \frac{\ln t}{t} \mathrm{~d} t=\left.\frac{1}{2} \ln ^{2} t\right|_{1} ^{x} = \frac{1}{2} \ln^{2} x.
$$

注意：$F(x) \neq \int F^{\prime}(t) \mathrm{~d} t$.

方法 2：把不一样的凑成一样的

$$
f\left(\frac{1}{x}\right)=\int_{1}^{\frac{1}{x}} \frac{\ln t}{1+t} \mathrm{~d} t \Rightarrow
$$

为了将 $f(\frac{1}{x})$ 的上下限变得和 $f(x)$ 一样，所以，令 $t=\frac{1}{u}$, 则：

$$
u=\frac{1}{t}, \quad \mathrm{~d} t=\frac{-1}{u^{2}} \mathrm{~d} u \Rightarrow
$$

$$
t \in\left(1, \frac{1}{x}\right) \Rightarrow u \in(1, x) \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln \frac{1}{u}}{1+\frac{1}{u}}\left(\frac{-1}{u^{2}}\right) \mathrm{~d} u \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{-\ln u^{-1}}{u^{2}\left(\frac{u+1}{u}\right)} \mathrm{~d} u=\int_{1}^{x} \frac{\ln u}{u(u+1)} \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \ln u \cdot \frac{1}{u(u+1)} \mathrm{~d} u \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \ln u \cdot\left(\frac{1}{u}-\frac{1}{u+1}\right) \mathrm{~d} u \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln u}{u} \mathrm{~d} u-\int_{1}^{x} \frac{\ln u}{u+1} \mathrm{~d} u \Rightarrow
$$

$$
u=t \Rightarrow
$$

$$
f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln t}{t} \mathrm{~d} t-\int_{1}^{x} \frac{\ln t}{t+1} \mathrm{~d} t \Rightarrow
$$

$$
f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\ln t}{t+1} \mathrm{~d} t+\int_{1}^{x} \frac{\ln t}{t} \mathrm{~d} t-\int_{1}^{x} \frac{\ln t}{t+1} \mathrm{~d} t =
$$

$$
\int_{1}^{x} \frac{\ln t}{t} \mathrm{~d} t=\left.\frac{1}{2} \ln { }^{2} t\right|_{1} ^{x}=\frac{1}{2} \ln ^{2} x
$$

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