Question

Find the following integrals.
$\int \frac{x^3+3x+4}{\sqrt{x}}dx.$

Answer 1

$\int \frac{x^3+3x+4}{\sqrt{x}}dx=\int \frac{x^3}{\sqrt{x}}dx+3\int \frac{x}{\sqrt{x}}dx+4\int \frac{1}{\sqrt{x}}dx=\int x^3\times x^{-\frac{1}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{\frac{-}{1}2}dx=\int x^{\frac{(6-1)}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{-\frac{1}{2}}dx=\int x^{\frac{5}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{-\frac{1}{2}}dx=\frac{x^{\frac{5}{2}+1}}{\left(\frac{5}{2}\right)+1}+\frac{3x^{\left(\frac{1}{2}\right)+1}}{\left(\frac{1}{2}\right)+1}+\frac{4x^{(-\frac{1}{2})+1}}{(-\frac{1}{2})+1}+C(\therefore \int x^{n}dx=\frac{x^{n+1}}{n+1})=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{4x^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{2}{7}{x}^{\frac{7}{2}}+2.x^{\frac{3}{2}}+8x^{\frac{1}{2}}+C$