# 积分式子中相似的部分越多越容易计算，但有时候需要我们拨开“云雾”

## 二、解析

$$\begin{cases} (x + 3) – \textcolor{springgreen}{4} = x – 1 \\ (5 – x) – \textcolor{springgreen}{4} = 1 – x \end{cases}$$

$$\begin{cases} (x + 1) – \textcolor{yellow}{2} = x – 1 \\ (3 – x) – \textcolor{yellow}{2} = 1 – x \end{cases}$$

\begin{aligned} I & = \\ \\ & \int_{1}^{+\infty} \frac{1}{e ^{x+3} + e ^{5-x}} \mathrm{~d} x \\ \\ & = \int_{1}^{+\infty} \frac{1}{ \textcolor{springgreen}{e ^{4} } \left( e ^{x-1} + e ^{1-x} \right) } \mathrm{~d} x \\ \\ & = \textcolor{springgreen}{ \frac{1}{e ^{4}} } \int_{1}^{+\infty} \frac{1}{ \left( e ^{x-1} + e ^{1-x} \right) } \mathrm{~d} x \\ \\ & \xlongequal[k \in (0, + \infty)]{k = x – 1} \frac{1}{e ^{4}} \int_{0}^{+\infty} \frac{1}{e ^{k} + e ^{-k}} \mathrm{~d} k \\ \\ & = \frac{1}{e ^{4}} \int_{0}^{+ \infty} \frac{1}{\frac{e ^{2k}}{e ^{k}} + \frac{1}{e ^{k}}} \mathrm{~d} x \\ \\ & = \frac{1}{e ^{4}} \int_{0}^{+\infty} \frac{e ^{k}}{e ^{2k} + 1} \mathrm{~d} x \\ \\ & = \frac{1}{e ^{4}} \int_{0}^{+\infty} \frac{1}{ 1 + (e ^{k}) ^{2}} \mathrm{~d} (e ^{k}) \\ \\ & = \frac{1}{e ^{4}} \arctan (e ^{k}) \Big|_{\textcolor{yellow}{\colorbox{magenta}{0}}} ^{+\infty} \\ \\ & = \frac{1}{e ^{4}} \times \left( \frac{\pi}{2} – \arctan \textcolor{yellow}{\colorbox{magenta}{1}} \right) \\ \\ & = \frac{1}{e ^{4}} \times \left( \frac{\pi}{2} – \frac{\pi}{4} \right) \\ \\ & = \frac{1}{e ^{4}} \times \frac{\pi}{4} \\ \\ & = \textcolor{springgreen}{\boldsymbol{\frac{\pi}{4 e ^{4}}}} \end{aligned}