Dùng phương pháp đổi biến số, tìm nguyên hàm của các hàm số sau:
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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