Đáp án:
Câu 1: D.
Câu 2: A.
Câu 3: B.
Câu 4: B.
Giải thích các bước giải:
$\begin{array}{l}
Cau\,1:\\
\int {2x\left( {1 + 3{x^3}} \right)dx} = \int {\left( {2x + 6{x^4}} \right)dx} \\
= \int {2xdx} + \int {6{x^4}dx} = {x^2} + 6.\frac{{{x^5}}}{5} + C = {x^2}\left( {1 + \frac{{6{x^3}}}{5}} \right) + C\\
Cau\,2:\\
\int {\left( {\frac{1}{{{x^2}}} - {x^2} - \frac{1}{3}} \right)dx} = \int {\frac{1}{{{x^2}}}dx} - \int {{x^2}dx} - \int {\frac{1}{3}dx} \\
= - \frac{1}{x} - \frac{{{x^3}}}{3} - \frac{1}{3}x + C = - \frac{{3 + {x^4} + {x^2}}}{{3x}} + C\\
Cau\,3:\\
\int {\sqrt[3]{x}d} = \int {{x^{\frac{1}{3}}}dx} = \frac{{{x^{\frac{1}{3} + 1}}}}{{\frac{1}{3} + 1}} + C = \frac{3}{4}{x^{\frac{4}{3}}} + C = \frac{{3\sqrt[3]{{{x^4}}}}}{4} + C = \frac{{3x\sqrt[3]{x}}}{4} + C\\
Cau\,4:\\
\int {\frac{1}{{x\sqrt x }}dx} = \int {{x^{ - \frac{3}{2}}}dx} = \frac{{{x^{ - \frac{3}{2} + 1}}}}{{ - \frac{3}{2} + 1}} + C\\
= \frac{{{x^{ - \frac{1}{2}}}}}{{ - \frac{1}{2}}} + C = - 2{x^{ - \frac{1}{2}}} + C = - \frac{2}{{\sqrt x }} + C
\end{array}$
