It is well known that $\mathrm{Ai}(x)$ and $\mathrm{Bi}(x)$ are two linear-independent solutions of Airy's ODE: $$y''-xy=0$$
Using Fourier transform, I am able to find out the integral form of $\mathrm{Ai}(x)$, but I would like to proceed further to find out the general solution for Airy's ODE. Hence, I substitute $y=f(x)\mathrm{Ai(x)}$ into Airy's ODE and obtain:
$$ \underbrace{f(x)\mathrm{Ai}''(x)}_M+2f'(x)\mathrm{Ai}'(x)+f''(x)\mathrm{Ai}(x)-xf(x)\mathrm{Ai}(x)=0 $$
Given that $\mathrm{Ai}''(x)=x\mathrm{Ai}(x)$, I am able to eliminate term $M$ and simplify the equation into
$$ 2f'(x)\mathrm{Ai}'(x)+f''(x)\mathrm{Ai}(x)=0 $$
Solving this reducible 2nd order ODE, I found that $f'(x)={C_1\over\mathrm{Ai}^2(x)}$, and this is where I am stuck: I was not able to calculate
$$\int{\mathrm{d}x\over\mathrm{Ai}^2(x)}$$
which is expected to be
$$ {\pi\mathrm{Bi}(x)\over\mathrm{Ai}(x)}+C $$
according to Wolfram Alpha, so I wonder if someone could help me find this integral so that I am able to obtain a general solution for Airy's ODE.
The Wronskian of the Airy functions is $$\frac{1}{\pi} = Ai(x) \, \frac{d \, Bi(x)}{dx} - Bi(x) \, \frac{d \, Ai(x)}{dx}$$ and leads to \begin{align} \int \frac{dx}{Ai^{2}(x)} &= \pi \, \int \frac{1}{\pi} \, \frac{dx}{Ai^{2}(x)} \\ &= \pi \, \int \left(Ai(x) \, \frac{d \, Bi(x)}{dx} - Bi(x) \, \frac{d \, Ai(x)}{dx} \right) \, \frac{dx}{Ai^{2}(x)} \\ &= \pi \, \int \frac{d}{dx} \left( \frac{Bi(x)}{Ai(x)} \right) \, dx \\ &= \frac{\pi \, Bi(x)}{Ai(x)} + c_{0}. \end{align}