giải giúp tôi

a) $\int(3\sqrt{x} + \frac{1}{\sqrt[3]{x}})dx$ Ta có: \[ \int 3\sqrt{x} dx = 3 \int x^{1/2} dx = 3 \cdot \frac{x^{3/2}}{\frac{3}{2}} = 2x^{3/2} \] \[ \int \frac{1}{\sqrt[3]{x}} dx = \int x^{-1/3} dx = \frac{x^{2/3}}{\frac{2}{3}} = \frac{3}{2} x^{2/3} \] Vậy: \[ \int(3\sqrt{x} + \frac{1}{\sqrt[3]{x}})dx = 2x^{3/2} + \frac{3}{2} x^{2/3} + C \] b) $\int \sqrt{x}(7x^2 - 3)dx$ (với $x > 0$) Ta có: \[ \int \sqrt{x}(7x^2 - 3)dx = \int (7x^{5/2} - 3x^{1/2}) dx \] \[ = 7 \int x^{5/2} dx - 3 \int x^{1/2} dx \] \[ = 7 \cdot \frac{x^{7/2}}{\frac{7}{2}} - 3 \cdot \frac{x^{3/2}}{\frac{3}{2}} \] \[ = 2x^{7/2} - 2x^{3/2} + C \] c) $\int \frac{(2x+1)^2}{x^2} dx$ Ta có: \[ \frac{(2x+1)^2}{x^2} = \frac{4x^2 + 4x + 1}{x^2} = 4 + \frac{4}{x} + \frac{1}{x^2} \] Vậy: \[ \int \left(4 + \frac{4}{x} + \frac{1}{x^2}\right) dx = 4x + 4 \ln|x| - \frac{1}{x} + C \] d) $\int (2^x + \frac{3}{x^2}) dx$ Ta có: \[ \int 2^x dx = \frac{2^x}{\ln 2} \] \[ \int \frac{3}{x^2} dx = 3 \int x^{-2} dx = 3 \cdot \frac{x^{-1}}{-1} = -\frac{3}{x} \] Vậy: \[ \int (2^x + \frac{3}{x^2}) dx = \frac{2^x}{\ln 2} - \frac{3}{x} + C \] Đáp số: a) $2x^{3/2} + \frac{3}{2} x^{2/3} + C$ b) $2x^{7/2} - 2x^{3/2} + C$ c) $4x + 4 \ln|x| - \frac{1}{x} + C$ d) $\frac{2^x}{\ln 2} - \frac{3}{x} + C$