Tính nguyên hàm √x - 2√x^2 + 1/ √x

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Tính: ∫x- 2x23 + 1x4dx

\[x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].

\[\frac{4}{5}x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].

\[x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].

\[\frac{4}{5}x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].

Giải thích

Chọn B

\[\int {\frac{{\sqrt x - 2\sqrt[3]{{{x^2}}} + 1}}{{\sqrt[4]{x}}}dx} = \int {\frac{{{x^{\frac{1}{2}}} - 2{x^{\frac{2}{3}}} + 1}}{{{x^{\frac{1}{4}}}}}dx = } \int {\left( {\frac{{{x^{\frac{1}{2}}}}}{{{x^{\frac{1}{4}}}}} - 2\frac{{{x^{\frac{2}{3}}}}}{{{x^{\frac{1}{4}}}}} + \frac{1}{{{x^{\frac{1}{4}}}}}} \right)dx} \]

\[ = \int {\left( {{x^{\frac{1}{4}}} - 2{x^{\frac{5}{{12}}}} + {x^{ - \frac{1}{4}}}} \right)dx = \frac{4}{5}} x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\]