Question
Find the following integrals in Exercises:

$\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$

 

Answer

$\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\text{x}^\frac{1}{2}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\bigg(3\text{x}^\frac{5}{2}+2\text{x}^{\frac{3}{2}}+3\text{x}^\frac{1}{2}\bigg)\text{ dx}$
$=\int3\text{x}^{\frac{5}{2}}\text{ dx }+\int2\text{x}^{\frac{3}{2}}\text{ dx}+\int3\text{x}^{\frac{1}{2}}\text{ dx}$
$=3\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+2 \frac { \text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+3 \frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c} $
$=3\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+2 \frac { \text{x}^{\frac{5}{2}}}{\frac{5}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{c} $
$\frac{6}{7}\text{x}^\frac{7}{2 }+ \frac{4}{5}\text{x}^\frac{5}{2 }+ 2\text{x}^\frac{3}{2}+\text{c} $

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