Tìm: a) nguyên hàm của (3x+4) . căn x; b) nguyên hàm của (2x+3) ^ 2/căn x

a) \(\int {\left( {3x + 4} \right)\sqrt[3]{x}} dx\) = \(\int {\left( {3x\sqrt[3]{x} + 4\sqrt[3]{x}} \right)} dx\)

                               = \(\int {3x\sqrt[3]{x}dx + \int {4\sqrt[3]{x}dx} } \)

                               = \(\int {3{x^{\frac{4}{3}}}dx} \) + \(\int {4{x^{\frac{1}{3}}}} dx\)

                               = \(\frac{9}{7}{x^2}\sqrt[3]{x} + 3x\sqrt[3]{x}\) + C

                               = \(\left( {\frac{9}{7}{x^2} + 3x} \right)\sqrt[3]{x}\) + C

b) \(\int {\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }}dx} \) = \(\int {\frac{{4{x^2} + 12x + 9}}{{\sqrt x }}dx} \)

                            = \(\int {\left( {4x\sqrt x + 12\sqrt x + \frac{9}{{\sqrt x }}} \right)dx} \)

                            = \(\int {4x\sqrt x dx + \int {12\sqrt x } } dx + \int {\frac{9}{{\sqrt x }}dx} \)

                            = \(4\int {{x^{\frac{3}{2}}}} dx + 12\int {{x^{\frac{1}{2}}}dx + 9\int {{x^{\frac{{ - 1}}{2}}}dx} } \)

                            = \(\frac{8}{5}{x^2}\sqrt x + 8x\sqrt x + 18\sqrt x + C\)

                            = \(\left( {\frac{8}{5}{x^2} + 8x + 18} \right)\sqrt x \) + C.