If int frac{5 cot x+1}{(cot x-1)(cot x-2) sin | vedclass

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(A) Let $I = \int \frac{5 \cot x+1}{(\cot x-1)(\cot x-2) \sin ^2 x} d x$. Substitute $\cot x = t$,then $-\operatorname{cosec}^2 x d x = d t$,which imp

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$\left(\cot x-1,(\cot x-2)^{-1}\right)$

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(A) Let $I = \int \frac{5 \cot x+1}{(\cot x-1)(\cot x-2) \sin ^2 x} d x$. Substitute $\cot x = t$,then $-\operatorname{cosec}^2 x d x = d t$,which implies $\operatorname{cosec}^2 x d x = -d t$. Thus,$I = -\int \frac{5t+1}{(t-1)(t-2)} dt$. Using partial fractions,$\frac{5t+1}{(t-1)(t-2)} = \frac{A}{t-1} + \frac{B}{t-2}$. $5t+1 = A(t-2) + B(t-1) = (A+B)t - (2A+B)$. Comparing coefficients,$A+B = 5$ and $2A+B = -1$. Solving these,$A = -6$ and $B = 11$. Therefore,$I = -\int \left( \frac{-6}{t-1} + \frac{11}{t-2} \right) dt = 6 \int \frac{dt}{t-1} - 11 \int \frac{dt}{t-2}$. $I = 6 \log |t-1| - 11 \log |t-2| + c = 6 \log |\cot x - 1| + 11 \log |(\cot x - 2)^{-1}| + c$. Comparing with $6 \log |f(x)| + 11 \log |g(x)| + c$,we get $f(x) = \cot x - 1$ and $g(x) = (\cot x - 2)^{-1}$. Thus,$(f(x), g(x)) = (\cot x - 1, (\cot x - 2)^{-1})$.

If $\int \frac{5 \cot x+1}{(\cot x-1)(\cot x-2) \sin ^2 x} d x = 6 \log |f(x)|+11 \log |g(x)|+c$,then $(f(x), g(x))=$

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