$\int \frac{1}{x^{2}(x-1)} d x \quad = ?$

Solve: ধরি, $\frac{1}{x^{2}(x-1)} \equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}$$\Rightarrow 1=A x(x-1)+B(x-1)+C x^{2} \cdots(1)$

(1) এ $x=0$ বসিয়ে পাই, $1=-B \Rightarrow B=-1$

(1) $এ x=1$ বসিয়ে পাই, 1= $\mathrm{C} \Rightarrow \mathrm{C}=1$

(1) এর উভয়পক্ষ থেকে $x^{2}$ এর সহগ সমীকৃত করে পাই, $0=\mathrm{A}+\mathrm{C} \Rightarrow \mathrm{A}=-\mathrm{C}=-1$

$\begin{array}{l} \therefore \quad \int \frac{1}{x^{2}(x-1)} d x=\int\left\{-\frac{1}{\dot{x}}-\frac{1}{x^{2}}+\frac{1}{x-1}\right\} d z \\ =-\ln |x|-\left(-\frac{1}{x}\right)+\ln |x-1|+c \\ =\ln \left|\frac{x-1}{x}\right|+\frac{1}{x}+c \end{array}$