Đề bài

a) \(\int {\left( {3x + 4} \right)\sqrt[3]{x}} dx\); b) \(\int {\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }}} dx\).

Lời giải chi tiết

a) Ta có \(\left( {3x + 4} \right)\sqrt[3]{x} = 3x\sqrt[3]{x} + 4\sqrt[3]{x} = 3{x^{\frac{4}{3}}} + 4{x^{\frac{1}{3}}}\). Do đó \(\int {\left( {3x + 4} \right)\sqrt[3]{x}} dx = \int {\left( {3{x^{\frac{4}{3}}} + 4{x^{\frac{1}{3}}}} \right)dx = } 3\int {{x^{\frac{4}{3}}}dx + } 4\int {{x^{\frac{1}{3}}}dx} \) \( = 3\frac{{{x^{\frac{7}{3}}}}}{{\left( {\frac{7}{3}} \right)}} + 4\frac{{{x^{\frac{4}{3}}}}}{{\left( {\frac{4}{3}} \right)}} + C = \frac{9}{7}{x^2}\sqrt[3]{x} + 3x\sqrt[3]{x} + C = \left( {\frac{9}{7}{x^2} + 3x} \right)\sqrt[3]{x} + C.\) b) Ta có \(\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }} = \frac{{4{x^2} + 12x + 9}}{{\sqrt x }} = 4x\sqrt x  + 12\sqrt x  + \frac{9}{{\sqrt x }} = 4{x^{\frac{3}{2}}} + 12{x^{\frac{1}{2}}} + \frac{9}{{\sqrt x }}\). Do đó \(\int {\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }}} dx = \int {\left( {4{x^{\frac{3}{2}}} + 12{x^{\frac{1}{2}}} + \frac{9}{{\sqrt x }}} \right)dx = } 4\int {{x^{\frac{3}{2}}}dx + } 12\int {{x^{\frac{1}{2}}}dx}  + 9\int {\frac{1}{{\sqrt x }}dx} \) \( = 4 \cdot \frac{{{x^{\frac{5}{2}}}}}{{\left( {\frac{5}{2}} \right)}} + 12 \cdot \frac{{{x^{\frac{3}{2}}}}}{{\left( {\frac{3}{2}} \right)}} + 9 \cdot 2\sqrt x  + C = \frac{8}{5}{x^2}\sqrt x  + 8x\sqrt x  + 18\sqrt x  + C = \left( {\frac{8}{5}{x^2} + 8x + 18} \right)\sqrt x  + C.\)