evaluate the following integral\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x \nfind the partial…

evaluate the following integral\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x \nfind the partial fraction decomposition of the integrand\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x = int left( - \frac { 1 } { x } - \frac { 16 } { x ^ { 2 } } + \frac { 1 } { x - 16 } \right) d x \nevaluate the indefinite integral\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x = square \n(use parentheses to clearly denote the argument of each functio

evaluate the following integral\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x \nfind the partial fraction decomposition of the integrand\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x = int left( - \frac { 1 } { x } - \frac { 16 } { x ^ { 2 } } + \frac { 1 } { x - 16 } \right) d x \nevaluate the indefinite integral\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x = square \n(use parentheses to clearly denote the argument of each functio

Answer

Explanation:

Step1: Integrate each term separately

$$ \begin{align*} \int\left(-\frac{1}{x}-\frac{16}{x^{2}}+\frac{1}{x - 16}\right)dx&=-\int\frac{1}{x}dx-16\int x^{-2}dx+\int\frac{1}{x - 16}dx \end{align*} $$

Step2: Apply integration formulas

  • For $\int\frac{1}{x}dx=\ln|x|+C_1$ (using the formula $\int\frac{1}{u}du=\ln|u|+C$ with $u = x$)
  • For $\int x^{-2}dx=\frac{x^{-2 + 1}}{-2+1}+C_2=\frac{x^{-1}}{-1}+C_2=-\frac{1}{x}+C_2$ (using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C,n\neq - 1$)
  • For $\int\frac{1}{x - 16}dx=\ln|x - 16|+C_3$ (using the formula $\int\frac{1}{u}du=\ln|u|+C$ with $u=x - 16$)

Step3: Combine the results

$$ \begin{align*} -\int\frac{1}{x}dx-16\int x^{-2}dx+\int\frac{1}{x - 16}dx&=-\ln|x|-16\left(-\frac{1}{x}\right)+\ln|x - 16|+C\ &=-\ln|x|+\frac{16}{x}+\ln|x - 16|+C \end{align*} $$

Answer:

$-\ln|x|+\frac{16}{x}+\ln|x - 16|+C$