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

Answer

$\int\frac{\text{x}^3+3\text{x}+4}{\sqrt{\text{x}}}\text{ dx}$
$=\int\Bigg(\frac{\text{x}^3}{\text{x}^{\frac{1}{2}}}+\frac{3\text{x}}{\text{x}^{\frac{1}{2}}}+\frac{4}{\text{x}^{\frac{1}{2}}}\Bigg)\text{ dx} $
$=\int\bigg(\text{x}^{3-\frac{1}{2}}+3\text{x}^{1-\frac {1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\bigg(\text{x}^{\frac{5}{2}}+3\text{x}^{\frac{1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\text{x}^\frac{5}{2}\text{ dx}+3\int\text{x}^\frac{1}{2}\text{ dx}+4\int\text{x}^\frac{-1}{2}\text{ dx} $
$=\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+4 \frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{c} $
$=\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+4 \frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{c} $
$=\frac{2}{7}\text{x}^{\frac{7}{2}}+2\text{x}^{\frac{3}{2}}+8\text{x}^{\frac{1}{2}}+\text{c}$

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