Bài tập 3.4 trang 164 SBT Toán 12

Bài tập 3.4 trang 164 SBT Toán 12

a) \(\mathop \smallint \nolimits {x^2}\sqrt[3]{{1 + {x^3}}}dx\) với x > - 1$\mathrm{}$

Đặt \(t = 1 + {x^3} \Rightarrow dt = 3{x^2}dx\)

 hay \({x^2}dx = \frac{{dt}}{3}\)

Ta có: 

\(\mathop \smallint \nolimits {x^2}\sqrt[3]{{1 + {x^3}}}dx = \frac{1}{3}\mathop \smallint \nolimits \sqrt[3]{t}dt = \frac{1}{3}.\frac{3}{4}{t^{\frac{4}{3}}} + C = \frac{1}{4}{(1 + {x^3})^{\frac{4}{3}}} + C\)

b) \(\mathop \smallint \nolimits x{e^{ - {x^2}}}dx\) 

Đặt \(t = {x^2} \Rightarrow dt = 2xdx\)

\(\mathop \smallint \nolimits x{e^{ - {x^2}}}dx = \mathop \smallint \nolimits {e^{ - t}}\frac{{dt}}{2} =  - \frac{{{e^{ - t}}}}{2} + C =  - \frac{{{e^{ - {x^2}}}}}{2} + C\)

c)  \(\mathop \smallint \nolimits \frac{x}{{{{\left( {1 + {x^2}} \right)}^2}}}dx\)

Đặt \(t = 1 + {x^2} \Rightarrow dt = 2xdx\)

\(\mathop \smallint \nolimits \frac{x}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = \mathop \smallint \nolimits \frac{{dt}}{{2{t^2}}} =  - \frac{1}{{2t}} + C =  - \frac{1}{{2(1 + {x^2})}} + C\)

d) \(\mathop \smallint \nolimits \frac{1}{{\left( {1 - x} \right)\sqrt x }}dx\)

Đặt \(t = \sqrt x  \Rightarrow dt = \frac{{dx}}{{2\sqrt x }}\)

\(\mathop \smallint \nolimits \frac{1}{{\left( {1 - x} \right)\sqrt x }}dx = \mathop \smallint \nolimits \frac{{2dt}}{{1 - {t^2}}} = \mathop \smallint \nolimits \left( {\frac{1}{{1 - t}} + \frac{1}{{1 + t}}} \right)dt =  - ln|1 - t| + ln|1 + t| + C\)

\( = ln\left| {\frac{{1 + t}}{{1 - t}}} \right| + C = ln\left| {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}} \right| + C\)

e) \(\int {\sin } \frac{1}{x}.\frac{1}{{{x^2}}}dx\)

Đặt \(t = \frac{1}{x} \Rightarrow dt =  - \frac{{dx}}{{{x^2}}}\)

\({\rm{ }}\int {\sin } \frac{1}{x}.\frac{1}{{{x^2}}}dx =  - {\rm{ }}\int {\sin } t.dt = \cos t + C = \cos \frac{1}{x} + C\)

g) \(\mathop \smallint \nolimits \frac{{{{\left( {\ln x} \right)}^2}}}{x}dx\)

Đặt \(t = lnx \Rightarrow dt = \frac{{dx}}{x}\)

\(\mathop \smallint \nolimits \frac{{{{\left( {\ln x} \right)}^2}}}{x}dx = \mathop \smallint \nolimits {t^2}dt = \frac{{{t^3}}}{3} + C = \frac{{{{(lnx)}^3}}}{3} + C\)

h) \(\int {\frac{{\sin x}}{{\sqrt[3]{{{{\cos }^2}x}}}}} dx\)

Đặt \(t = \cos x \Rightarrow dt =  - \sin xdx\)

\(\int {\frac{{\sin x}}{{\sqrt[3]{{{{\cos }^2}x}}}}} dx = \int {\frac{{ - dt}}{{{t^{\frac{2}{3}}}}}}  =  - 3{t^{\frac{1}{3}}} + C =  - 3\sqrt[3]{{\cos x}} + C\)

-- Mod Toán 12