kaoyan2advanced 高等数学 第125题

**答案**：B **解析**： 步骤1：$\displaystyle f(x)=\int_1^x \frac{\ln(1+t)}{t}\mathrm{d}t$，则$\displaystyle f\left(\frac{1}{x}\right)=\int_1^{1/x} \frac{\ln(1+t)}{t}\mathrm{d}t$。 步骤2：令$u=1/t$，则$\displaystyle \int_1^{1/x} \frac{\ln(1+t)}{t}\mathrm{d}t = \int_1^x \frac{\ln(1+1/u)}{1/u} \cdot (-\frac{1}{u^2})\mathrm{d}u = \int_1^x \frac{\ln\frac{u+1}{u}}{u}\mathrm{d}u = \int_1^x \frac{\ln(1+u)}{u}\mathrm{d}u - \int_1^x \frac{\ln u}{u}\mathrm{d}u$。 步骤3：故$\displaystyle f(x)+f\left(\frac{1}{x}\right) = 2\int_1^x \frac{\ln(1+t)}{t}\mathrm{d}t - \int_1^x \frac{\ln t}{t}\mathrm{d}t$。 步骤4：当$x=2$时，$\displaystyle \int_1^2 \frac{\ln t}{t}\mathrm{d}t = \frac{1}{2}\ln^2 2$，而$\displaystyle 2\int_1^2 \frac{\ln(1+t)}{t}\mathrm{d}t$需计算，但注意到$f(x)+f(1/x)$可化简：由步骤2，$\displaystyle f(1/x)=\int_1^x \frac{\ln(1+t)}{t}\mathrm{d}t - \int_1^x \frac{\ln t}{t}\mathrm{d}t$，所以$\displaystyle f(x)+f(1/x)=2\int_1^x \frac{\ln(1+t)}{t}\mathrm{d}t - \int_1^x \frac{\ln t}{t}\mathrm{d}t$。 步骤5：另一种方法：$\displaystyle f(x)+f(1/x)=\int_1^x \frac{\ln(1+t)}{t}\mathrm{d}t + \int_1^{1/x} \frac{\ln(1+t)}{t}\mathrm{d}t$，令$t=1/u$于第二个积分，得$\displaystyle \int_1^x \frac{\ln(1+1/u)}{u}\mathrm{d}u = \int_1^x \frac{\ln(1+u)}{u}\mathrm{d}u - \int_1^x \frac{\ln u}{u}\mathrm{d}u$，相加得$\displaystyle 2\int_1^x \frac{\ln(1+u)}{u}\mathrm{d}u - \int_1^x \frac{\ln u}{u}\mathrm{d}u$。 步骤6：当$x=2$时，$\displaystyle \int_1^2 \frac{\ln u}{u}\mathrm{d}u = \frac{1}{2}\ln^2 2$，而$\displaystyle \int_1^2 \frac{\ln(1+u)}{u}\mathrm{d}u$无法直接化简，但注意到$f(2)+f(1/2)$可特殊计算：$\displaystyle f(2)=\int_1^2 \frac{\ln(1+t)}{t}\mathrm{d}t$，$\displaystyle f(1/2)=\int_1^{1/2} \frac{\ln(1+t)}{t}\mathrm{d}t = -\int_{1/2}^1 \frac{\ln(1+t)}{t}\mathrm{d}t$，相加得$\displaystyle \int_1^2 \frac{\ln(1+t)}{t}\mathrm{d}t - \int_{1/2}^1 \frac{\ln(1+t)}{t}\mathrm{d}t = \int_{1/2}^2 \frac{\ln(1+t)}{t}\mathrm{d}t - \int_{1/2}^1 \frac{\ln(1+t)}{t}\mathrm{d}t$，不易直接得。 步骤7：利用恒等式：$\displaystyle f(x)+f(1/x)=\frac{1}{2}\ln^2 x$（常见结论），验证：对$x>0$，求导得$\displaystyle f'(x)+f'(1/x)\cdot(-1/x^2)=\frac{\ln(1+x)}{x} + \frac{\ln(1+1/x)}{1/x}\cdot(-1/x^2)=\frac{\ln(1+x)}{x} - \frac{\ln(1+1/x)}{x} = \frac{\ln x}{x}$，积分得$\displaystyle \frac{1}{2}\ln^2 x + C$，取$x=1$得$f(1)+f(1)=0$，故$C=0$。所以$\displaystyle f(2)+f(1/2)=\frac{1}{2}\ln^2 2$。 **难度**：★★★★☆