evaluate the indefinite integral.\n int\frac{9}{x^{3}-3x^{2}}dx=square

Explanation:

Step1: Factor the denominator

First, factor out $x^{2}$ from the denominator: $x^{3}-3x^{2}=x^{2}(x - 3)$. Then, use partial - fraction decomposition. Let $\frac{9}{x^{2}(x - 3)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x - 3}$. [9=Ax(x - 3)+B(x - 3)+Cx^{2}] If $x = 0$, then $9=B(-3)$, so $B=- 3$. If $x = 3$, then $9=C\times3^{2}$, so $C = 1$. Expand the right - hand side: $9=Ax^{2}-3Ax + Bx-3B + Cx^{2}=(A + C)x^{2}+(-3A + B)x-3B$. Since $B=-3$ and $C = 1$, and the coefficient of $x^{2}$ is $0$ ($A + C=0$), then $A=-1$. So, $\frac{9}{x^{2}(x - 3)}=-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}$.

Step2: Integrate term - by - term

(\int\frac{9}{x^{2}(x - 3)}dx=\int\left(-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}\right)dx) We know that (\int\frac{1}{x}dx=\ln|x|+C_1), (\int x^{-n}dx=\frac{x^{-n + 1}}{-n+1}+C_2) ((n\neq1)) and (\int\frac{1}{u}du=\ln|u|+C_3). (\int\left(-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}\right)dx=-\int\frac{1}{x}dx-3\int x^{-2}dx+\int\frac{1}{x - 3}dx) (=-\ln|x|-3\times\frac{x^{-2 + 1}}{-2 + 1}+\ln|x - 3|+C) (=-\ln|x|+\frac{3}{x}+\ln|x - 3|+C) (=\ln\left|\frac{x - 3}{x}\right|+\frac{3}{x}+C)

Answer:

(\ln\left|\frac{x - 3}{x}\right|+\frac{3}{x}+C)