Giải bài 4.3 trang 7

Đề bài

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

b) $\int \frac{{\left(2x+3\right)}^2}{\sqrt{x}}dx$.

Lời giải chi tiết

a) Ta có $\left(3x+4\right)\sqrt[3]{3x}=3x\sqrt[3]{3x}+4\sqrt[3]{3x}=3x^{\frac{4}{3}}+4x^{\frac{1}{3}}$.

Do đó $\int \left(3x+4\right)\sqrt[3]{3x}dx=\int \left(3x^{\frac{4}{3}}+4x^{\frac{1}{3}}\right)dx=3\int x^{\frac{4}{3}}dx+4\int x^{\frac{1}{3}}dx$

$=3\frac{x^{\frac{7}{3}}}{\left(\frac{7}{3}\right)}+4\frac{x^{\frac{4}{3}}}{\left(\frac{4}{3}\right)}+C=\frac{9}{7}{x}^2\sqrt[3]{3x}+3x\sqrt[3]{3x}+C=\left(\frac{9}{7}{x}^2+3x\right)\sqrt[3]{3x}+C.$

b) Ta có $\frac{{\left(2x+3\right)}^2}{\sqrt{x}}=\frac{4x^2+12x+9}{\sqrt{x}}=4x\sqrt{x}+12\sqrt{x}+\frac{9}{\sqrt{x}}=4x^{\frac{3}{2}}+12x^{\frac{1}{2}}+\frac{9}{\sqrt{x}}$.

Do đó $\int \frac{{\left(2x+3\right)}^2}{\sqrt{x}}dx=\int \left(4x^{\frac{3}{2}}+12x^{\frac{1}{2}}+\frac{9}{\sqrt{x}}\right)dx=4\int x^{\frac{3}{2}}dx+12\int x^{\frac{1}{2}}dx+9\int \frac{1}{\sqrt{x}}dx$

$=4\cdot \frac{x^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}+12\cdot \frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+9\cdot 2\sqrt{x}+C=\frac{8}{5}{x}^2\sqrt{x}+8x\sqrt{x}+18\sqrt{x}+C=\left(\frac{8}{5}{x}^2+8x+18\right)\sqrt{x}+C.$