Giải bài 24 trang 162 SGK giải tích nâng cao 12

a) Đặt ${{x}^{3}}=t\Rightarrow 3{{x}^{2}}dx=dt$

$\Rightarrow \int\limits_{1}^{2}{{{x}^{2}}{{e}^{{{x}^{3}}}}dx}=\dfrac{1}{3}\int\limits_{1}^{8}{{{e}^{t}}dt}=\dfrac{1}{3}{{e}^{t}}\left| _{\begin{smallmatrix} \\ \\\\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 8 \\ \\\\ \end{smallmatrix}} \right.=\dfrac{1}{3}\left( {{e}^{8}}-e \right)$

b) Đặt $\ln x=t\Rightarrow \dfrac{1}{x}dx=dt$

$\Rightarrow \int\limits_{1}^{3}{\dfrac{1}{x}{{\left( \ln x \right)}^{2}}dx}=\int\limits_{0}^{\ln 3}{{{t}^{2}}dt}=\dfrac{1}{3}{{t}^{3}}\left| _{\begin{smallmatrix} \\ \\\\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \ln 3 \\\\\\ \end{smallmatrix}}=\dfrac{{{\left( \ln 3 \right)}^{3}}}{3} \right.$

c) Đặt $\sqrt{1+{{x}^{2}}}=t\Rightarrow 1+{{x}^{2}}={{t}^{2}}\Rightarrow xdx=tdt$

$\Rightarrow \int\limits_{0}^{\sqrt{3}}{x\sqrt{1+{{x}^{2}}}dx}=\int\limits_{1}^{2}{t^2dt}=\dfrac{1}{3}{{t}^{3}}\left| _{\begin{smallmatrix} \\ \\\\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \\\\ \end{smallmatrix}} \right.=\dfrac{1}{3}\left( 8-1 \right)=\dfrac{7}{3}$

d) Đặt $3{{x}^{3}}=t\Rightarrow 9{{x}^{2}}dx=dt$

$\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}{{e}^{3{{x}^{3}}}}dx}=\dfrac{1}{9}\int\limits_{0}^{3}{{{e}^{t}}dt}=\dfrac{1}{9}{{e}^{t}}\left| _{\begin{smallmatrix} \\ \\\\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 3 \\ \\\\ \end{smallmatrix}} \right.=\dfrac{1}{9}\left( {{e}^{3}}-1 \right)$

e) Đặt $1+\sin x=t\Rightarrow \cos xdx=dt$

$\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{\dfrac{\cos x}{1+\sin x}dx}=\int\limits_{1}^{2}{\dfrac{dt}{t}}=\ln \left| t \right|\left| _{\begin{smallmatrix} \\ \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \\ \end{smallmatrix}} \right.=\ln 2$