evaluate the following integral.\n int\frac{9}{x^{3}-3x^{2}}dx \nfind the partial - fraction decomposition…

Explanation:

Step1: Factor the denominator

$x^{3}-3x^{2}=x^{2}(x - 3)$

Step2: Set up partial - fraction decomposition

$\frac{9}{x^{2}(x - 3)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x - 3}$ $9=Ax(x - 3)+B(x - 3)+Cx^{2}$

Step3: Find the values of A, B, and C

If $x = 0$: $9=B(0 - 3)\Rightarrow B=- 3$ If $x = 3$: $9=C\times3^{2}\Rightarrow C = 1$ Expand $9=Ax(x - 3)+B(x - 3)+Cx^{2}$: $9=Ax^{2}-3Ax + Bx-3B + Cx^{2}$ $9=(A + C)x^{2}+(-3A + B)x-3B$ Since $B=-3$ and $C = 1$, and the coefficient of $x^{2}$ is $0$ ($A + C=0$), then $A=-1$ So $\frac{9}{x^{2}(x - 3)}=-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}$ $\int\frac{9}{x^{3}-3x^{2}}dx=\int\left(-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}\right)dx$

Step4: Integrate term - by - term

$\int\left(-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}\right)dx=-\int\frac{1}{x}dx-3\int x^{-2}dx+\int\frac{1}{x - 3}dx$ $=-\ln|x|+\frac{3}{x}+\ln|x - 3|+C$

Answer:

$\int\frac{9}{x^{3}-3x^{2}}dx=\int\left(-\frac{1}{x}-\frac{3}{x^{2}}+\frac{1}{x - 3}\right)dx$; $-\ln|x|+\frac{3}{x}+\ln|x - 3|+C$