$$ \begin{aligned} & \text { If } \int \frac{(x+3)}{(x-1)^2(2 x-1)} d x \\ & =\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+k, \text { then } A+B+C= \end{aligned} $$

If $\int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x=f(x) \cdot \cos (g(x))+c$, then $f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=$

Let $x \neq \frac{-3}{5}, \frac{2}{5}$, if $f\left(\frac{2 x+1}{5 x+3}\right)=x+2$, then $\int f(x) d x=$

If $\int e^x \cos x d x=\frac{e^x}{2}(\cos x+\sin x)$ and

$$ \int \frac{\cos \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)}{(3-2 x)^2} d x=\frac{f(x)}{24}[\cos (g(x))+\sin (g(x))]+c $$

then $g(1)=$