Bài 2 Trang 141 SGK Đại số và Giải tích 12 Nâng cao

LG a

\(\int {\left( {\sqrt x  + \root 3 \of x } \right)dx;} \)

Lời giải chi tiết:

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

LG b

\(\int {{{x\sqrt x  + \sqrt x } \over {{x^2}}}} dx;\)

Lời giải chi tiết:

\(\int {\dfrac{{x\sqrt x  + \sqrt x }}{{{x^2}}}dx} \)\( = \int {\dfrac{{x.{x^{\dfrac{1}{2}}} + {x^{\dfrac{1}{2}}}}}{{{x^2}}}dx} \) \( = \int {\dfrac{{{x^{\dfrac{3}{2}}} + {x^{\dfrac{1}{2}}}}}{{{x^2}}}dx} \) \( = \int {\left( {\dfrac{{{x^{\dfrac{3}{2}}}}}{{{x^2}}} + \dfrac{{{x^{\dfrac{1}{2}}}}}{{{x^2}}}} \right)dx} \) \( = \int {\left( {{x^{ - \dfrac{1}{2}}} + {x^{ - \dfrac{3}{2}}}} \right)dx} \) \( = \dfrac{{{x^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + \dfrac{{{x^{ - \dfrac{3}{2} + 1}}}}{{ - \dfrac{3}{2} + 1}} + C\) \( = \dfrac{{{x^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + \dfrac{{{x^{ - \dfrac{1}{2}}}}}{{ - \dfrac{1}{2}}} + C\) \( = 2{x^{\dfrac{1}{2}}} - 2{x^{ - \dfrac{1}{2}}} + C\)  \( = 2\sqrt x  - \dfrac{2}{{\sqrt x }} + C\)

LG c

\(\int {4{{\sin }^2}xdx;} \) 

Lời giải chi tiết:

\(\int {4{{\sin }^2}xdx} \) \( = \int {4.\dfrac{{1 - \cos 2x}}{2}dx} \) \( = \int {2\left( {1 - \cos 2x} \right)dx} \) \( = 2\int {\left( {1 - \cos 2x} \right)dx} \) \( = 2\left( {\int {dx}  - \int {\cos 2xdx} } \right)\) \( = 2\left( {x - \dfrac{{\sin 2x}}{2}} \right) + C\) \( = 2x - \sin 2x + C\)

LG d

\(\int {{{1 + \cos 4x} \over 2}dx.} \)

Lời giải chi tiết:

\(\int {\dfrac{{1 + \cos 4x}}{2}dx} \)\( = \dfrac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} \)  \( = \dfrac{1}{2}\left( {x + \dfrac{{\sin 4x}}{4}} \right) + C\) \( = \dfrac{x}{2} + \dfrac{{\sin 4x}}{8} + C\)

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