# If ∫[log (log x) + 1 / (log x)2] dx = x [f (x) - g (x)] + c

1) f (x) = log (log x); g (x) = 1 / log x

2) f (x) = log x, g (x) = 1 / log x

3) f (x) = 1 / [x log x]; g (x) = log (log x)

4) f (x) = 1 / [x log x], g (x) = 1 / log x

Solution: (1) f (x) = log (log x); g (x) = 1 / log x

$$\begin{array}{l}\begin{array}{l} \int\left[\ln (\ln x)+\frac{1}{(\ln x)^{2}}\right] d x \\ \Rightarrow \int \ln (\ln x) d x+\int \frac{d x}{(\ln x)^{2}} \\ \Rightarrow x \ln (\ln x)-\int \frac{1}{\ln x} \frac{1}{x} x d x+\int \frac{d x}{(\ln x)^{2}} \\ \Rightarrow x \ln (\ln x)-\int \frac{d x}{\ln x}+\int \frac{d x}{(\ln x)^{2}} \\ \Rightarrow x \ln (\ln x)-\frac{1}{\ln x} x-\int \frac{d x}{(\ln x)^{2}}+\int \frac{d x}{(\ln x)^{2}} \\ \Rightarrow x \ln (\ln x)-\frac{x}{\ln x}+c \\ \Rightarrow x\left[\ln (\ln x)-\frac{1}{\ln x}\right]+c \\ \therefore \text { Compare: } f(x)=\ln (\ln x) \\ g(x)=\frac{1}{\ln x} \end{array}\end{array}$$