武忠祥老师每日一题||不定积分基础训练（六）

解法一：
$$求出f(x),进而对f(x)进行积分。$$
$$令\ln x=t,原式f(t)=\frac{\ln (1+e^t)}{e^t}$$
$$\begin{gathered} 则\int f(x)\mathrm{d}x=\int \frac{\ln (1+e^x)}{e^x}\mathrm{d}x \\ =\int \ln (1+e^x)e^{-x}\mathrm{d}x \end{gathered}$$
$$=-\int \ln (1+e^x)\mathrm{d}{e}^{-x}$$
$$=-\ln (1+e^x)e^{-x}+\int e^{-x}\mathrm{d}\ln (1+e^x)$$
$$=-\ln (1+e^x)e^{-x}+\int e^{-x}\frac{1}{1+e^x}\times e^x\mathrm{d}x$$
$$=-\ln (1+e^x)e^{-x}+\int \frac{1}{1+e^x}\mathrm{d}x$$
$$=-\ln (1+e^x)e^{-x}+x-\ln (e^x+1)+C$$

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$$计算\int \frac{1}{1+e^x}\mathrm{d}x:$$
$$原式=\int \frac{(1+e^x)-e^x}{1+e^x}\mathrm{d}x$$
$$=\int (1-\frac{e^x}{1+e^x})\mathrm{d}x$$
$$=x-\int \frac{\mathrm{d}(e^x+1)}{1+e^x}$$
$$=x-\ln (e^x+1)+C$$

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解法二：
$$令t=\ln x(x=e^t)\int f(\ln x)\mathrm{d}\ln x$$
$$=\int \frac{\ln (1+x)}{x}\times \frac{1}{x}\mathrm{d}x$$
$$=\int \frac{\ln (1+x)}{x^2}\mathrm{d}x$$
$$=-\int \ln (1+x)\mathrm{d}\frac{1}{x}$$
$$=-\frac{\ln (1+x)}{x}+\int \frac{1}{x}\times \frac{1}{x+1}\mathrm{d}x$$
$$=-\frac{\ln (1+x)}{x}+\ln |\frac{x}{x+1}|+C$$
$$=-\frac{\ln (1+e^t)}{e^t}+\ln |\frac{e^t}{e^t+1}|+C$$
由于积分变量为x，则所求为
$$-\frac{\ln (1+e^x)}{e^x}+\ln |\frac{e^x}{e^x+1}|+C$$