Họ nguyên hàm của hàm số \(f\left(x\right)=\dfrac{x+\sqrt{x}+1}{\sqrt[3]{x}}\) là
\(\dfrac{5}{3}x^{\frac{5}{3}}+\dfrac{7}{6}x^{\frac{7}{6}}+\dfrac{2}{3}x^{\frac{2}{3}}+C\).\(\dfrac{3}{5}x^{\frac{5}{3}}+\dfrac{6}{7}x^{\frac{7}{6}}+\dfrac{3}{2}x^{\frac{2}{3}}+C\).\(\dfrac{2}{3}x^{-\frac{1}{3}}+\dfrac{1}{6}x^{\frac{-5}{6}}-\dfrac{1}{3}x^{\frac{-4}{3}}+C\).\(-3x^{-\frac{1}{3}}-\dfrac{6}{5}x^{\frac{-5}{6}}-\dfrac{3}{4}x^{\frac{-4}{3}}+C\).Hướng dẫn giải:\(f\left(x\right)=\dfrac{x+\sqrt{x}+1}{\sqrt[3]{x}}=\dfrac{x+x^{\frac{1}{2}}+1}{x^{\frac{1}{3}}}\)
\(=x^{\frac{2}{3}}+x^{\frac{1}{6}}+x^{-\frac{1}{3}}\)
Suy ra \(\int f\left(x\right)dx=\int x^{\frac{2}{3}}dx+\int x^{\frac{1}{6}}dx+\int x^{-\frac{1}{3}}dx\)
\(=\frac{1}{\frac{2}{3}+1}x^{\frac{2}{3}+1}+\frac{1}{\frac{1}{6}+1}x^{\frac{1}{6}+1}+\frac{1}{-\frac{1}{3}+1}x^{-\frac{1}{3}+1}+C\)
\(=\frac{3}{5}x^{\frac{5}{3}}+\frac{6}{7}x^{\frac{7}{6}}+\frac{3}{2}x^{\frac{2}{3}}+C\)
