Prove
Question:

$\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$

Solution:

$\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$

$=\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x$

$=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{3}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+\mathrm{C}$

$=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+\mathrm{C}$

$=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+\mathrm{C}$

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