求解定积分 $\int_{-2}^{2}$ $x \ln(1+e^{x})$ $\mathrm{d} x$

一、题目

$$
\int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x = ?
$$

难度评级：

二、解析

本题可以借助区间再现公式求解，同时，在求解本题时需要格外注意正负号。

令 $t$ $=$ $2$ $+$ $(-2)$ $-$ $x$, 于是：

$$
t = -x
$$

$$
x = -t
$$

$$
\mathrm{d} x = – \mathrm{d} t
$$

$$
x \in (-2, 2) \Rightarrow t \in (2, -2)
$$

Next

于是：

$$
\int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x =
$$

$$
– \int_{2}^{-2} (-t) \ln(1+e^{-t}) \mathrm{d} t =
$$

$$
– \int_{-2}^{2} t \ln(1+e^{-t}) \mathrm{d} t.
$$

Next

若令 $t$ $=$ $x$, 则：

$$
– \int_{-2}^{2} x \ln(1+e^{-x}) \mathrm{d} x =
$$

$$
– \int_{-2}^{2} x \ln(1+\frac{1}{e^{x}}) \mathrm{d} x =
$$

$$
– \int_{-2}^{2} x \ln(\frac{e^{x} + 1}{e^{x}}) \mathrm{d} x =
$$

$$
– \int_{-2}^{2} x \Big[ \ln(1 + e^{x}) – \ln e^{x} \Big] \mathrm{d} x =
$$

$$
– \int_{-2}^{2} x \ln(1 + e^{x}) \mathrm{d} x + \int_{-2}^{2} x\ln e^{x} \mathrm{d} x =
$$

$$
– \int_{-2}^{2} x \ln(1 + e^{x}) \mathrm{d} x + \int_{-2}^{2} x^{2} \mathrm{d} x.
$$

Next

进而：

$$
\int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x = – \int_{-2}^{2} x \ln(1 + e^{x}) \mathrm{d} x + \int_{-2}^{2} x^{2} \mathrm{d} x \Rightarrow
$$

$$
2 \int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x = \int_{-2}^{2} x^{2} \mathrm{d} x \Rightarrow
$$

$$
\int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x = \frac{1}{2} \int_{-2}^{2} x^{2} \mathrm{d} x \Rightarrow
$$

$$
\int_{-2}^{2} x \ln(1+e^{x}) \mathrm{d} x = \frac{1}{2} \cdot \frac{1}{3} x^{3} \Big|_{-2}^{2} = \frac{1}{6} (8 + 8) = \frac{8}{3}.
$$

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