分母上的根号可以通过求导去除

一、题目

$$
\int \frac{x e^{x}}{\sqrt{1 + e^{x}}} \mathrm{d} x = ?
$$

难度评级：

二、解析

解法 1：通过求导凑出分母有理化

由于：

$$
[ (1 + e^{x})^{\frac{1}{2}} ]^{\prime} = \frac{1}{2} (1 + e^{x})^{\frac{-1}{2}} \cdot \textcolor{orange}{e^{x}} \Rightarrow
$$

$$
[ (1 + e^{x})^{\frac{1}{2}} ]^{\prime} = \frac{1}{2} \frac{e^{x}}{\sqrt{1 + e^{x}}}.
$$

注 意 ：$[ (1 + e^{x})^{\frac{1}{2}} ]^{\prime}$ $\neq$ $\frac{1}{2} (1 + e^{x})^{\frac{-1}{2}}$

Next

所以：

$$
\int \frac{x e^{x}}{\sqrt{1 + e^{x}}} \mathrm{~d} x =
$$

$$
2 \int x \mathrm{~d} (\sqrt{1 + e^{x}}) =
$$

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

Next

若令 $t = \sqrt{1 + e^{x}}$, 则：

$$
t^{2} = 1 + e^{x} \Rightarrow
$$

$$
e^{x} = t^{2} – 1 \Rightarrow
$$

$$
\ln(t^{2} – 1) = x.
$$

Next

于是：

$$
\int \sqrt{1 + e^{x}} \mathrm{~d} x =
$$

$$
\int t \mathrm{~d} [ \ln(t^{2} – 1) ] =
$$

$$
\int t \cdot \frac{2 t}{t^{2} – 1} \mathrm{~d} t =
$$

$$
\int \frac{2 t^{2}}{t^{2} – 1} \mathrm{~d} t =
$$

$$
2 \int \frac{t^{2} – 1 + 1}{t^{2} – 1} \mathrm{~d} t =
$$

$$
2 \int \Big( 1 + \frac{1}{t^{2} – 1} \Big) \mathrm{~d} t =
$$

$$
2 \textcolor{orange}{ \int 1 \mathrm{~d} t } + 2 \int \frac{1}{t^{2} – 1} \mathrm{~d} t =
$$

$$
2 \textcolor{orange}{t} + 2 \int \frac{1}{(t + 1) (t – 1)} \mathrm{~d} t =
$$

$$
2 t + 2 \cdot (\frac{-1}{2}) \int \Big( \frac{1}{t + 1} – \frac{1}{t – 1} \Big) \mathrm{~d} t =
$$

$$
2 t – \big( \ln |t + 1| – \ln |t – 1| \big) + C =
$$

$$
2 t – \ln \Big| \frac{t+1}{t-1} \Big| + C.
$$

Next

令 $t = \sqrt{1 + e^{x}}$ $\Rightarrow$

$$
2 t – \ln \Big| \frac{t+1}{t-1} \Big| + C \Rightarrow
$$

$$
\int \sqrt{1 + e^{x}} \mathrm{~d} x = 2 \sqrt{1 + e^{x}} – \ln \Big| \frac{\sqrt{1 + e^{x}} + 1}{\sqrt{1 + e^{x}} – 1} \Big| + C.
$$

Next

于是：

$$
2 x \sqrt{1 + e^{x}} – 2 \int \sqrt{1 + e^{x}} \mathrm{~d} x =
$$

$$
2 x \sqrt{1 + e^{x}} – 2 \Big[ 2 \sqrt{1 + e^{x}} – \ln \Big| \frac{\sqrt{1 + e^{x}} + 1}{\sqrt{1 + e^{x}} – 1} \Big| \Big] + C =
$$

$$
2 x \sqrt{1 + e^{x}} – 4 \sqrt{1 + e^{x}} + 2 \ln \Big| \frac{\sqrt{1 + e^{x}} + 1}{\sqrt{1 + e^{x}} – 1} \Big| + C.
$$

Next

或者：

$$
2 x \sqrt{1 + e^{x}} – 4 \sqrt{1 + e^{x}} – 2 \ln \Big| \frac{\sqrt{1 + e^{x}} – 1}{\sqrt{1 + e^{x}} + 1} \Big| + C.
$$

解法 2：对根号部分整体代换

令 $t = \sqrt{ 1 + e^{x} }$, 则：

$$
e^{x} = t^{2} – 1
$$

$$
x = \ln (t^{2} – 1)
$$

Next

于是：

$$
\int \frac{x e^{x}}{ \sqrt{1 + e^{x}} } \mathrm{d} x =
$$

$$
\int \frac{ \ln (t^{2} – 1) (t^{2} – 1)}{ t } \mathrm{d} [\ln (t^{2} – 1)] =
$$

$$
\int \frac{ \ln (t^{2} – 1) (t^{2} – 1)}{ t } \cdot \frac{2t}{t^{2} – 1} \mathrm{d} t =
$$

$$
2 \int \ln (t^{2} – 1) \mathrm{d} t \Rightarrow
$$

Next

分部积分 $\Rightarrow$

$$
\textcolor{red}{2} \Big[ t \ln (t^{2} – 1) – \int t \mathrm{d} \ln (t^{2} – 1) \Big] =
$$

$$
\textcolor{red}{2} t \ln (t^{2} – 1) – \textcolor{red}{2} \int \frac{2t^{2}}{t^{2} – 1} \mathrm{d} t =
$$

注 意 ：去括号的时候，一定不要忘记把括号外面的数字乘进去，同时在去括号的时候要考虑一下括号内的加减号是否需要变号。

$$
2 t \ln (t^{2} – 1) – 4 \int \frac{t^{2} – 1 + 1}{t^{2} – 1} \mathrm{d} t =
$$

$$
2 t \ln (t^{2} – 1) – 4 \Big[ \int \frac{t^{2} – 1}{t^{2} – 1} \mathrm{d} t + \int \frac{1}{t^{2} – 1} \mathrm{d} t \Big] =
$$

$$
2 t \ln (t^{2} – 1) – 4 \Big[ \int 1 \mathrm{d} t + \int \frac{1}{t^{2} – 1} \mathrm{d} t \Big] =
$$

$$
2 t \ln (t^{2} – 1) – 4 t – 4 \int \frac{1}{(t+1)(t-1)} \mathrm{d} t =
$$

$$
2 t \ln (t^{2} – 1) – 4 t – 4 \cdot (\frac{-1}{2}) \int \Big[ \frac{1}{t+1} – \frac{1}{t-1} \Big] \mathrm{d} t =
$$

$$
2 t \ln (t^{2} – 1) – 4 t + 2 \Big[ \ln|t+1| – \ln|t-1| \Big] + C =
$$

$$
2 t \ln (t^{2} – 1) – 4 t + 2 \ln \Big|\frac{t+1}{t-1} \Big| + C.
$$

Next

令 $t = \sqrt{ 1 + e^{x} }$, 则：

$$
2 t \ln (t^{2} – 1) – 4 t + 2 \ln \Big|\frac{t+1}{t-1} \Big| + C =
$$

$$
2 x \sqrt{ 1 + e^{x} } – 4 \sqrt{ 1 + e^{x} } + 2 \ln \Big|\frac{\sqrt{ 1 + e^{x} } + 1}{\sqrt{ 1 + e^{x} } – 1} \Big| + C.
$$

其中，$C$ 为任意常数。

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