Bài 5 Trang 145 SGK Đại số và Giải tích 12 Nâng cao

LG a

\(f\left( x \right) = {{9{x^2}} \over {\sqrt {1 - {x^3}} }}\)

Lời giải chi tiết:

Đặt \(\sqrt {1 - {x^3}}  = u\) \( \Rightarrow {u^2} = 1 - {x^3}\) \( \Rightarrow 2udu =  - 3{x^2}dx\)

\( \Rightarrow \int {f\left( x \right)dx} \)\( = \int {\dfrac{{ - 3.\left( { - 3{x^2}} \right)dx}}{{\sqrt {1 - {x^3}} }}} \) \( = \int {\dfrac{{ - 3.2udu}}{u}} \)  \( =  - 6\int {du}  =  - 6u + C\) \( =  - 6\sqrt {1 - {x^3}}  + C\)

Cách khác:

Đặt \(1 - {x^3} = u \Rightarrow du =  - 3{x^2}dx\)

\( \Rightarrow \int {f\left( x \right)dx} \)\( = \int {\dfrac{{ - 3.\left( { - 3{x^2}dx} \right)}}{{\sqrt {1 - {x^3}} }}}  = \int {\dfrac{{ - 3du}}{{\sqrt u }}} \) \( = \int { - 3{u^{ - \dfrac{1}{2}}}du}  =  - 3.\dfrac{{{u^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C\) \( =  - 3.\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C =  - 6{u^{\dfrac{1}{2}}} + C\)  \( =  - 6\sqrt u  + C =  - 6\sqrt {1 - {x^3}}  + C\)

LG b

\(f\left( x \right) = {1 \over {\sqrt {5x + 4} }}\)

Lời giải chi tiết:

Đặt \(u = \sqrt {5x + 4}  \Rightarrow {u^2} = 5x + 4\) \( \Rightarrow 2udu = 5dx \Rightarrow dx = {{2u.du} \over 5}\)

\( \Rightarrow \int {f\left( x \right)dx}  = \int {\dfrac{1}{u}.\dfrac{{2udu}}{5}}  = \int {\dfrac{2}{5}du} \) \( = \dfrac{2}{5}u + C = \dfrac{2}{5}\sqrt {5x + 4}  + C\)

Cách 2:

\(\int {\dfrac{1}{{\sqrt {5x + 4} }}dx}  = \int {\dfrac{1}{5}.\dfrac{{d\left( {5x + 4} \right)}}{{{{\left( {5x + 4} \right)}^{\dfrac{1}{2}}}}}} \)\( = \int {\dfrac{1}{5}.{{\left( {5x + 4} \right)}^{ - \dfrac{1}{2}}}d\left( {5x + 4} \right)} \) \( = \dfrac{1}{5}.\dfrac{{{{\left( {5x + 4} \right)}^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C\) \( = \dfrac{1}{5}.\dfrac{{{{\left( {5x + 4} \right)}^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C\)  \( = \dfrac{2}{5}{\left( {5x + 4} \right)^{\dfrac{1}{2}}} + C\) \( = \dfrac{2}{5}\sqrt {5x + 4}  + C\)

Cách 3

Đặt \(5x + 4 = u\) \( \Rightarrow 5dx = du \Rightarrow dx = \dfrac{{du}}{5}\)

\( \Rightarrow \int {f\left( x \right)dx}  = \int {\dfrac{1}{{\sqrt u }}.\dfrac{{du}}{5}}  \) \(= \dfrac{2}{5}\int {\dfrac{1}{{2\sqrt u }}du} \) \( = \dfrac{2}{5}\sqrt u  + C = \dfrac{2}{5}\sqrt {5x + 4}  + C\)

LG c

\(f\left( x \right) = x\root 4 \of {1 - {x^2}} \)

Lời giải chi tiết:

Đặt \(u = \root 4 \of {1 - {x^2}}  \) \(\Rightarrow {u^4} = 1 - {x^2}\) \( \Rightarrow 4{u^3}du =  - 2xdx\) \(  \Rightarrow xdx =  - 2{u^3}du\)

\( \Rightarrow \int {f\left( x \right)dx} \)\( = \int { - 2{u^3}.udu}  =  - 2\int {{u^4}du} \) \( =  - 2.\dfrac{{{u^5}}}{5} + C =  - \dfrac{{2{u^5}}}{5} + C\) \( =  - \dfrac{{2{{\left( {\sqrt[4]{{1 - {x^2}}}} \right)}^5}}}{5} + C\)  \( =  - \dfrac{{2\left( {1 - {x^2}} \right)\sqrt[4]{{1 - {x^2}}}}}{5} + C\)

Cách khác:

Đặt \(1 - {x^2} = u\) \( \Rightarrow  - 2xdx = du \Rightarrow xdx =  - \dfrac{{du}}{2}\)

\( \Rightarrow \int {f\left( x \right)dx} \) \( = \int {\sqrt[4]{u}.\left( { - \dfrac{{du}}{2}} \right)} \) \( =  - \dfrac{1}{2}\int {{u^{\dfrac{1}{4}}}du} \) \( =  - \dfrac{1}{2}.\dfrac{{{u^{\dfrac{1}{4} + 1}}}}{{\dfrac{1}{4} + 1}} + C\)\( =  - \dfrac{1}{2}.\dfrac{{{u^{\dfrac{5}{4}}}}}{{\dfrac{5}{4}}} + C =  - \dfrac{2}{5}{u^{\dfrac{5}{4}}} + C\) \( =  - \dfrac{2}{5}\sqrt[4]{{{{\left( {1 - {x^2}} \right)}^5}}} + C\)  \( =  - \dfrac{2}{5}\left( {1 - {x^2}} \right)\sqrt[4]{{1 - {x^2}}} + C\)

LG d

\(f\left( x \right) = {1 \over {\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)

Lời giải chi tiết:

Đặt \(\displaystyle u = 1 + \sqrt x \) \(\displaystyle \Rightarrow du = {{du} \over {2\sqrt x }} \) \(\displaystyle \Rightarrow {{dx} \over {\sqrt x }} = 2du\)

\(\displaystyle \Rightarrow \int {{{dx} \over {\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}}  \) \(\displaystyle  = \int {{{2u} \over {{u^2}}}}  =  - {2 \over u} + C \) \(\displaystyle =  - {2 \over {1 + \sqrt x }} + C.\)

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