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Find the value of \(\rm \int {{x^3 +3x+4}\over {\sqrt x}} dx\)
Concept:Some useful formulas are:\(\rm ? x^n dx = \frac {(x^{n+1})} {(n+1)} +C; \ n?1\)Calculation:\(\rm \int {{x^3\; +\;3x\;+\;4}\over {\sqrt x}} dx\)\(\)= \(\rm \int \frac{x^3}{\sqrt x}dx + \int \frac{3x}{\sqrt x}dx + \int \frac{4}{\sqrt x}dx\)= \(\rm \int x^{5\over 2} dx + \int 3x^{1\over2}dx\ + \int 4 x^{-1\over2} dx\)= \(\rm \frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1} + 3\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+4\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C\), where C is the constant of integration= \(\rm{ 2\over 7}\times x^{7\over2} + {3\;\times \;2\over 3}x^{3\over2}+ 4\times2x^{1\over2}+C \)=\(\rm{ 2\over 7}\times x^{7\over2} + 2x^{3\over2}+ 8x^{1\over2}+C \)
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