Giải bài 8 trang 145 SGK giải tích nâng cao 12

a) $\int{{{x}^{2}}{{\left( \dfrac{{{x}^{3}}}{18}-1 \right)}^{5}}dx}$

Đặt $u=\dfrac{{{x}^{3}}}{18}-1\Leftrightarrow \dfrac{{{x}^{2}}}{6}dx=du\Leftrightarrow {{x}^{2}}dx=6du$

$\begin{aligned} \Rightarrow \int{{{x}^{2}}{{\left( \dfrac{{{x}^{3}}}{18}-1 \right)}^{5}}dx}&=6\int{{{u}^{5}}du}={{u}^{6}}+C \\ & ={{\left( \dfrac{{{x}^{3}}}{18}-1 \right)}^{6}}+C \\ \end{aligned}$

b) $\int{\dfrac{1}{{{x}^{2}}}\sin \dfrac{1}{x}\cos \dfrac{1}{x}dx}$

Đặt $u=\sin \dfrac{1}{x}\Leftrightarrow du=-\dfrac{1}{{{x}^{2}}}\cos \dfrac{1}{x}dx$

$\begin{aligned} \Rightarrow \int{\dfrac{1}{{{x}^{2}}}\sin \dfrac{1}{x}\cos \dfrac{1}{x}dx}&=-\int{udu}=-\dfrac{{{u}^{2}}}{2}+C \\ & =-\dfrac{1}{2}\sin \dfrac{1}{x}+C \\ \end{aligned}$

c) $\int{{{x}^{3}}{{e}^{x}}dx}$

Đặt $\left\{ \begin{aligned} & {{x}^{3}}=u \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & 3{{x}^{2}}dx=du \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\Rightarrow \int{{{x}^{3}}{{e}^{x}}dx}={{x}^{3}}{{e}^{x}}-3\int{{{x}^{2}}{{e}^{x}}dx}$

Đặt $\left\{ \begin{aligned} & {{x}^{2}}=u \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & 2xdx=du \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\Rightarrow \int{{{x}^{2}}{{e}^{x}}dx}={{x}^{2}}{{e}^{x}}-2\int{x{{e}^{x}}dx}$

Đặt $\left\{ \begin{aligned} & x=u \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & dx=du \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\Rightarrow \int{x{{e}^{x}}dx}=x{{e}^{x}}-\int{{{e}^{x}}dx}=x{{e}^{x}}-{{e}^{x}} \\ \begin{aligned} \Rightarrow \int{{{x}^{3}}{{e}^{x}}dx}&={{x}^{3}}{{e}^{x}}-3\left[ {{x}^{2}}{{e}^{x}}-2\left( x{{e}^{x}}-{{e}^{x}} \right) \right] \\ & =\left( {{x}^{3}}-3{{x}^{2}}+6x-6 \right){{e}^{x}}+C \\ \end{aligned}$

d) $\int{{{e}^{\sqrt{3x-9}}}dx}$

Đặt $\sqrt{3x-9}=t\Leftrightarrow 3x-9={{t}^{2}}\Leftrightarrow 3dx=2tdt$

$\Rightarrow \int{{{e}^{\sqrt{3x-9}}}dx}=\dfrac{2}{3}\int{t{{e}^{t}}dt}$

Đặt $\left\{ \begin{aligned} & u=t \\ & {{e}^{t}}dt=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=dt \\ & v={{e}^{t}} \\ \end{aligned} \right.$

$\Rightarrow \int{t{{e}^{t}}dt}=t.{{e}^{t}}-\int{{{e}^{t}}dt}=\left( t-1 \right).{{e}^{t}} \\ \begin{aligned} \Rightarrow \int{{{e}^{\sqrt{3x-9}}}dx}&=\dfrac{2}{3}\left( t-1 \right).{{e}^{t}}+C \\ & =\dfrac{2}{3}\left( \sqrt{3x-9}-1 \right){{.e}^{\sqrt{3x-9}}}+C \\ \end{aligned}$