题目
求不定积分：
$$
\int e^{2x} \arctan \sqrt{e^{x} – 1} \mathrm{d} x.
$$
解析
由题可得：
$$
\int e^{2x} \arctan \sqrt{e^{x} – 1} \mathrm{d} x \Rightarrow
$$
$$
\frac{1}{2} \int \arctan \sqrt{e^{x} – 1} \mathrm{d} (e^{2x}) \Rightarrow
$$
$$
{\color{Red}
\frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \int e^{2x} \mathrm{d} (\arctan \sqrt{e^{x}-1}) \Bigg]}.
$$
又：
$$
(\arctan \sqrt{e^{x}-1}) ^{‘} =
$$
$$
\frac{1}{1 + (\sqrt{e^{x} – 1})^{2}} \cdot (\sqrt{e^{x} – 1})^{‘} =
$$
$$
\frac{1}{e^{x}} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}} \cdot e^{x} =
$$
$$
\frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}}.
$$
于是：
$$
{\color{Red}
\frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \int e^{2x} \mathrm{d} (\arctan \sqrt{e^{x}-1}) \Bigg]} \Rightarrow
$$
$$
\frac{1}{2} \Bigg[ e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x \Bigg] \Rightarrow
$$
$$
{\color{Red}
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x}.
$$
又：
$$
(\sqrt{e^{x} – 1}) ^{‘} = \frac{1}{2} \cdot \frac{1}{\sqrt{e^{x} – 1}} \cdot e^{x} = \frac{1}{2} \frac{e^{x}}{\sqrt{e^{x} -1}}.
$$
于是：
$$
{\color{Red}
\int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x} \Rightarrow
$$
$$
\int e^{x} \cdot e^{x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x \Rightarrow
$$
$$
2 \int e^{x} \mathrm{d} (\sqrt{e^{x} – 1}) \Rightarrow
$$
$$
2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \int \sqrt{e^{x} – 1} \mathrm{d} (e^{x}) \Bigg] \Rightarrow
$$
$$
2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C.
$$
于是：
$$
{\color{Red}
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \int e^{2x} \cdot \frac{1}{\sqrt{e^{x} – 1}} \mathrm{d} x} \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{4} \cdot 2 \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ e^{x} \sqrt{e^{x} – 1} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ e^{x} (e^{x} – 1)^{\frac{1}{2}} – \frac{2}{3} (e^{x} – 1)^{\frac{3}{2}} \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{2} \Bigg[ (e^{x} – 1)^{\frac{1}{2}} [e^{x} – \frac{2}{3} (e^{x} – 1) ] \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \frac{1}{2} \cdot \Bigg[e^{x} – \frac{2}{3} (e^{x} – 1) \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{2} e^{x} – \frac{1}{3} (e^{x} – 1) \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{2} e^{x} – \frac{1}{3} e^{x} + \frac{1}{3} \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – (e^{x} – 1)^{\frac{1}{2}} \cdot \Bigg[ \frac{1}{6} e^{x} + \frac{1}{3} \Bigg] + C \Rightarrow
$$
$$
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \sqrt{e^{x} – 1} \cdot \frac{1}{6} ( e^{x} + 2) + C \Rightarrow
$$
$$
{\color{Red}
\frac{1}{2} e^{2x} \arctan \sqrt{e^{x} – 1} – \frac{1}{6} ( e^{x} + 2) \sqrt{e^{x} – 1} + C}.
$$