5. *\n\\(\\int(\\frac{8}{x}-\\frac{5}{x^{2}}+\\frac{6}{x^{3}})dx\\)\n a. \\(8\\ln|x|-\\frac{5}{x}-\\frac{3}{x…

Explanation:

Step1: Split the integral

Use the sum - difference rule of integration $\int(f(x)\pm g(x)\pm h(x))dx=\int f(x)dx\pm\int g(x)dx\pm\int h(x)dx$. So, $\int(\frac{8}{x}-\frac{5}{x^{2}}+\frac{6}{x^{3}})dx=\int\frac{8}{x}dx-\int\frac{5}{x^{2}}dx+\int\frac{6}{x^{3}}dx$.

Step2: Integrate each term

Recall the integration formulas: $\int\frac{1}{x}dx = \ln|x|+C$, $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int\frac{8}{x}dx$, since $\int\frac{1}{x}dx=\ln|x|+C$, then $\int\frac{8}{x}dx = 8\ln|x|+C_1$. For $\int\frac{5}{x^{2}}dx=\int5x^{-2}dx$, using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n + 1}+C$, we have $\int5x^{-2}dx=5\times\frac{x^{-2 + 1}}{-2+1}=- \frac{5}{x}+C_2$. For $\int\frac{6}{x^{3}}dx=\int6x^{-3}dx$, using the power - rule, $\int6x^{-3}dx=6\times\frac{x^{-3 + 1}}{-3 + 1}=-\frac{3}{x^{2}}+C_3$.

Step3: Combine the results

$\int(\frac{8}{x}-\frac{5}{x^{2}}+\frac{6}{x^{3}})dx=8\ln|x|-\frac{5}{x}-\frac{3}{x^{2}}+C$ (where $C = C_1+C_2+C_3$).

Answer:

A. $8\ln|x|-\frac{5}{x}-\frac{3}{x^{2}}+C$