题目

设函数 $f(x)$ 在 $(\dfrac{1}{2}, +\infty)$ 上可导，且 $\lim\limits_{h\to0}\dfrac{f[(x+h)^2] - f(x^2+h)}{h}=1,f(1)=1$，求$f(x)$

解答

$$
\begin{aligned}
&
\lim\limits_{h\to0}\dfrac{f[(x+h)^2] - f(x^2+h)}{h}
\\
=&
\lim\limits_{h\to0}\dfrac{f[(x+h)^2] - f(x^2) + f(x^2) - f(x^2+h)}{h}
\\
=&
\lim\limits_{h\to0}\dfrac{f[(x+h)^2] - f(x^2)}{h} -
\lim\limits_{h\to0}\dfrac{f(x^2+h)-f(x^2)}{h}
\\
=&
[f(x^2)]’ - f’(x^2) = f’(x^2) \cdot (2x - 1)
\end{aligned}
$$

令 $f’(x^2) \cdot (2x-1) = 1$，换元可得 $f’(x) = \dfrac{1}{2\sqrt{x}-1}$，左右取积分：

$$
\begin{aligned}
\int f’(x) dx &= \int \dfrac{1}{2\sqrt{x}-1} dx \\
f(x) &= \int \dfrac{2u - 1 + 1}{2u-1} du = \int (1 + \dfrac{1}{2u-1}) du \\
f(x) &= u + \dfrac{1}{2}\ln(2u - 1) + C \\
f(x) &= \sqrt{x} + \dfrac{1}{2}\ln(2\sqrt{x} - 1) + C \\
\end{aligned}
$$

代入初值：$f(1) = 1 + C = 1$，故 $C = 0$，于是有：

$$
f(x) = \sqrt{x} + \dfrac{1}{2}\ln(2\sqrt{x} - 1)
$$