Question

Find the integral: $\int \frac{x^3+3x+4}{\sqrt{x}}dx$

Answer 1

$\int \frac{x^3+3x+4}{\sqrt{x}}dx$
$=\int (\frac{x^3}{\sqrt{x}}+\frac{3x}{\sqrt{x}}+\frac{4}{\sqrt{x}})dx$
$=\int [x^{(3-\frac{1}{2})}+3x^{(1-\frac{1}{2})}+4x^{(-\frac{1}{2})}]dx$
$=\int (x^{\frac{5}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}})dx$
$=\int x^{\frac{5}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{-\frac{1}{2}}dx$
$=\left(\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right)+3\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)+4\left(\frac{x^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right)+C$
$=\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+4\left(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$
$=\frac{2x^{\frac{7}{2}}}{7}+2x^{\frac{3}{2}}+8\sqrt{x}+C$

Where $C$ is constant of integration