Solution. (a) First write the root as a power, since the rule only applies to powers: \[\sqrt {x}=x^{\frac {1}{2}}.\] Then \[\int \left (3x^2+4x^{\frac {1}{2}}\right )dx =\frac {3x^{3}}{3}+\frac {4x^{\frac {3}{2}}}{\frac {3}{2}}+c =x^3+\frac {8}{3}x^{\frac {3}{2}}+c.\] Dividing by \(\frac {3}{2}\) is multiplying by \(\frac {2}{3}\), so \(4\) becomes \(4\times \frac {2}{3}=\frac {8}{3}\).

(b) Again convert first: \[\frac {1}{\sqrt {x}}=x^{-\frac {1}{2}}.\] Adding one to \(-\frac {1}{2}\) gives \(\frac {1}{2}\): \[\int \left (x^{-\frac {1}{2}}+5\right )dx=\frac {x^{\frac {1}{2}}}{\frac {1}{2}}+5x+c =2\sqrt {x}+5x+c.\]

Solution. (a) Rewrite both terms as powers: \[\frac {3}{x^2}=3x^{-2},\qquad 2\sqrt {x}=2x^{\frac {1}{2}}.\] \[\int \left (3x^{-2}-2x^{\frac {1}{2}}\right )dx =\frac {3x^{-1}}{-1}-\frac {2x^{\frac {3}{2}}}{\frac {3}{2}}+c =-\frac {3}{x}-\frac {4}{3}x^{\frac {3}{2}}+c.\]

(b) Multiply out before integrating: \[x\left (x^2+\frac {3}{x}\right )=x^3+3.\] The second term simplifies to a constant, which is easy to miss: \[\int \left (x^3+3\right )dx=\frac {x^4}{4}+3x+c.\]

(c) Divide each term of the numerator by \(\sqrt {x}=x^{\frac {1}{2}}\), subtracting indices: \[\frac {x^3+2x}{x^{\frac {1}{2}}}=x^{3-\frac {1}{2}}+2x^{1-\frac {1}{2}} =x^{\frac {5}{2}}+2x^{\frac {1}{2}}.\] Now integrate: \[\int \left (x^{\frac {5}{2}}+2x^{\frac {1}{2}}\right )dx =\frac {x^{\frac {7}{2}}}{\frac {7}{2}}+\frac {2x^{\frac {3}{2}}}{\frac {3}{2}}+c =\frac {2}{7}x^{\frac {7}{2}}+\frac {4}{3}x^{\frac {3}{2}}+c.\]