(int\frac{dx}{x(x - 1)^2})

Explanation:

Step1: Use partial - fraction decomposition

Let $\frac{1}{x(x - 1)^2}=\frac{A}{x}+\frac{B}{x - 1}+\frac{C}{(x - 1)^2}$. Then $1=A(x - 1)^2+Bx(x - 1)+Cx$. Set $x = 0$, we get $1=A(0 - 1)^2$, so $A = 1$. Set $x=1$, we get $1 = C\times1$, so $C = 1$. Expand $A(x - 1)^2+Bx(x - 1)+Cx$: $A(x^2-2x + 1)+B(x^2-x)+Cx=(A + B)x^2+(-2A - B + C)x+A$. Since $A = 1$ and $C = 1$, and the coefficient of $x^2$ is $0$ ($A + B=0$), then $B=-1$. So $\frac{1}{x(x - 1)^2}=\frac{1}{x}-\frac{1}{x - 1}+\frac{1}{(x - 1)^2}$.

Step2: Integrate term - by - term

$\int\frac{1}{x(x - 1)^2}dx=\int\frac{1}{x}dx-\int\frac{1}{x - 1}dx+\int\frac{1}{(x - 1)^2}dx$. We know that $\int\frac{1}{x}dx=\ln|x|+C_1$, $\int\frac{1}{x - 1}dx=\ln|x - 1|+C_2$, and $\int\frac{1}{(x - 1)^2}dx=-\frac{1}{x - 1}+C_3$.

Answer:

$\ln|x|-\ln|x - 1|-\frac{1}{x - 1}+C$