1. ∫(x³ - 4x⁵ + 9 - √x + 2/∛x + 3eˣ + 1/x)dx

Explanation:

Step1: Apply sum - difference rule of integration

$\int (x^{3}-4x^{5}+9 - \sqrt{x}+\frac{2}{\sqrt[3]{x}}+3e^{x}+\frac{1}{x})dx=\int x^{3}dx-4\int x^{5}dx + 9\int dx-\int x^{\frac{1}{2}}dx+2\int x^{-\frac{1}{3}}dx+3\int e^{x}dx+\int\frac{1}{x}dx$

Step2: Use power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$ and $\int e^{x}dx=e^{x}+C$, $\int\frac{1}{x}dx=\ln|x|+C$

For $\int x^{3}dx=\frac{x^{3 + 1}}{3+1}=\frac{x^{4}}{4}$; for $-4\int x^{5}dx=-4\times\frac{x^{5+1}}{5 + 1}=-\frac{4x^{6}}{6}=-\frac{2x^{6}}{3}$; for $9\int dx=9x$; for $-\int x^{\frac{1}{2}}dx=-\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=-\frac{2}{3}x^{\frac{3}{2}}$; for $2\int x^{-\frac{1}{3}}dx=2\times\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}=2\times\frac{x^{\frac{2}{3}}}{\frac{2}{3}} = 3x^{\frac{2}{3}}$; for $3\int e^{x}dx=3e^{x}$; for $\int\frac{1}{x}dx=\ln|x|$

Step3: Combine the results

$\frac{x^{4}}{4}-\frac{2x^{6}}{3}+9x-\frac{2}{3}x^{\frac{3}{2}}+3x^{\frac{2}{3}}+3e^{x}+\ln|x|+C$

Answer:

$\frac{x^{4}}{4}-\frac{2x^{6}}{3}+9x-\frac{2}{3}x^{\frac{3}{2}}+3x^{\frac{2}{3}}+3e^{x}+\ln|x|+C$