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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Aicharya · Accepted Answer

$I=\int \frac{5 \cot x+1}{(\cot x-1)(\cot x-2) \sin ^2 x} d x$ Let $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$, so $\begin{aligned} & I=-\int \frac{5 t+1}{(t-1)(t-2)} d t \ & \text {Let } \frac{5 t+1}{(t-1)(t-2)}=\frac{A}{t-1}+\frac{B}{t-2} \ & \Rightarrow \quad 5 t+1=(A+B) t-(2 A+B) \ & \text {So, } \quad A+B=5 \text { and } 2 A+B=-1 \ & \Rightarrow \quad A=-6 \text { and } B=11 \ & \therefore \quad I=6 \int \frac{d t}{t-1}-11 \int \frac{d t}{t-2} \ & =6 \log |t-1|-11 \log |t-2|+c \ & =6 \log |\cot x-1|+11 \log \left|(\cot x-2)^{-1}\right|+c \end{aligned}$ On comparing, we get $(f(x), g(x))=\left((\cot x-1),(\cot x-2)^{-1}\right)$ Hence, option (a) is correct.

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…

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