# 2018年考研数二第15题解析：分部积分法、求导

## 题目

$$\int e^{2x} \arctan \sqrt{e^{x} – 1} \mathrm{d} x.$$

## 解析

$$\int e^{2x} \arctan \sqrt{e^{x} – 1} \mathrm{d} x \Rightarrow$$

$$\frac{1}{2} \int \arctan \sqrt{e^{x} – 1} \mathrm{d} (e^{2x}) \Rightarrow$$

$${\color{Red} \frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \int e^{2x} \mathrm{d} (\arctan \sqrt{e^{x}-1}) \Bigg]}.$$

$$(\arctan \sqrt{e^{x}-1}) ^{‘} =$$

$$\frac{1}{1 + (\sqrt{e^{x} – 1})^{2}} \cdot (\sqrt{e^{x} – 1})^{‘} =$$

$$\frac{1}{e^{x}} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}} \cdot e^{x} =$$

$$\frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}}.$$

$${\color{Red} \frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \int e^{2x} \mathrm{d} (\arctan \sqrt{e^{x}-1}) \Bigg]} \Rightarrow$$

$$\frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x \Bigg] \Rightarrow$$

$${\color{Red} \frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x}.$$

$$(\sqrt{e^{x} – 1}) ^{‘} = \frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}} \cdot e^{x} = \frac{1}{2} \frac{e^{x}}{\sqrt{e^{x} -1}}.$$

$${\color{Red} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x} \Rightarrow$$

$$\int e^{x} \cdot e^{x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x \Rightarrow$$

$$2 \int e^{x} \mathrm{d} (\sqrt{e^{x} – 1}) \Rightarrow$$

$$2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \int \sqrt{e^{x} – 1} \mathrm{d} (e^{x}) \Bigg] \Rightarrow$$

$$2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C.$$

$${\color{Red} \frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x} \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \cdot 2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ e^{x} (e^{x} – 1)^{\frac{1}{2}} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ (e^{x} – 1)^{\frac{1}{2}} [e^{x} – \frac{2}{3} (e^{x} – 1) ] \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \frac{1}{2} \cdot \Bigg[e^{x} – \frac{2}{3} (e^{x} – 1) \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{2} e^{x} – \frac{1}{3} (e^{x} – 1) \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{2} e^{x} – \frac{1}{3} e^{x} + \frac{1}{3} \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{6} e^{x} + \frac{1}{3} \Bigg] + C \Rightarrow$$

$$\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \sqrt{e^{x} – 1} \cdot \frac{1}{6} ( e^{x} + 2) + C \Rightarrow$$

$${\color{Red} \frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{6} ( e^{x} + 2) \sqrt{e^{x} – 1} + C}.$$