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

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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:

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

SGK Toán 12 Nâng cao