evaluate the following integral.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x $$\n\nfind the…

evaluate the following integral.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x $$\n\nfind the partial fraction decomposition of the integrand.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x = int left( - \frac { 1 } { x } - \frac { 3 } { x ^ { 2 } } + \frac { 1 } { x - 3 } \right) d x $$\n\nevaluate the indefinite integral.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x = $$\n\n(use parentheses to clearly denote the argument of each function.)

evaluate the following integral.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x $$\n\nfind the partial fraction decomposition of the integrand.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x = int left( - \frac { 1 } { x } - \frac { 3 } { x ^ { 2 } } + \frac { 1 } { x - 3 } \right) d x $$\n\nevaluate the indefinite integral.\n\n$$ int \frac { 9 } { x ^ { 3 } - 3 x ^ { 2 } } d x = $$\n\n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Integrate term - by - term

We know that (\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).

Step2: Apply integral formulas

  • For (\int\frac{1}{x}dx=\ln|x|+C_1).
  • For (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), when (n=-2), (\int x^{-2}dx=\frac{x^{-2 + 1}}{-2+1}=-\frac{1}{x}+C_2).
  • For (\int\frac{1}{u}du=\ln|u|+C) (using substitution (u=x - 3), (du=dx)), (\int\frac{1}{x - 3}dx=\ln|x - 3|+C_3).

Step3: Combine the results

[ \begin{align*} -\int\frac{1}{x}dx-3\int x^{-2}dx+\int\frac{1}{x - 3}dx&=-\ln|x|-3\left(-\frac{1}{x}\right)+\ln|x - 3|+C\ &=-\ln|x|+\frac{3}{x}+\ln|x - 3|+C \end{align*} ]

Answer:

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