Giải bài 5 trang 127 – SGK môn Giải tích lớp 12

a) Đặt $\sqrt{1+x}=t\Rightarrow \left\{ \begin{aligned} & dt=\dfrac{1}{2\sqrt{1+x}}dx \\ & x={{t}^{2}}-1 \\ \end{aligned} \right.$

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

b) 

$\begin{aligned} \int\limits_{1}^{64}{\dfrac{1+\sqrt{x}}{\sqrt[3]{x}}}dx &=\int\limits_{1}^{64}{\left( {{x}^{-\frac{1}{3}}}+{{x}^{\frac{1}{6}}} \right)}dx \\ & =\left( \dfrac{3}{2}{{x}^{\frac{2}{3}}}+\dfrac{6}{7}{{x}^{\frac{7}{6}}} \right)\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 64 \\ \end{smallmatrix}} \right. \\ & =24+\dfrac{768}{7}-\dfrac{3}{2}-\dfrac{6}{7}\\&=\dfrac{1839}{14} \\ \end{aligned}$

c) Đặt $\left\{ \begin{aligned} & {{x}^{2}}=u \\ & {{e}^{3x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=2xdx \\ & v=\dfrac{1}{3}{{e}^{3x}} \\ \end{aligned} \right.$

$\begin{aligned} \int\limits_{0}^{2}{{{x}^{2}}{{e}^{3x}}dx}&=\dfrac{1}{3}{{x}^{2}}{{e}^{3x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right.-\dfrac{2}{3}\int\limits_{0}^{2}{x{{e}^{3x}}dx} \\ & =\dfrac{4}{3}{{e}^{6}}-\dfrac{2}{3}\int\limits_{0}^{2}{x{{e}^{3x}}dx} \\ \end{aligned}$

Đặt $\left\{ \begin{aligned} & x=u \\ & {{e}^{3x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=\dfrac{1}{3}{{e}^{3x}} \\ \end{aligned} \right.$

$\begin{aligned} \int\limits_{0}^{2}{x{{e}^{3x}}dx}&=\dfrac{1}{3}x{{e}^{3x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right.-\dfrac{1}{3}\int\limits_{0}^{2}{{{e}^{3x}}dx} \\ & =\dfrac{2}{3}{{e}^{6}}-\dfrac{1}{9}{{e}^{3x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right. \\ & =\dfrac{2}{3}{{e}^{6}}-\dfrac{1}{9}{{e}^{6}} +\dfrac{1}{9} \\ & =\dfrac{5}{9}{{e}^{6}} +\dfrac{1}{9} \\ \end{aligned}$

$\Rightarrow \int\limits_{0}^{2}{{{x}^{2}}{{e}^{3x}}dx}=\dfrac{4}{3}{{e}^{6}}-\dfrac{2}{3}.\left(\dfrac{5}{9}{{e}^{6}}+\dfrac{1}{9}\right)=\dfrac{26{{e}^{6}}}{27}-\dfrac{2}{27}$

d)

$\begin{aligned} \int\limits_{0}^{\pi }{\sqrt{1+\sin 2x}dx}&=\int\limits_{0}^{\pi }{\sqrt{{{\left( \operatorname{sinx}+cosx \right)}^{2}}}dx} \\ & =\int\limits_{0}^{\pi }{\left| \sin x+\cos x \right|dx} \\ & =\sqrt{2}\int\limits_{0}^{\pi }{\left| \sin \left( x+\dfrac{\pi }{4} \right) \right|dx} \\ & =\sqrt{2}\int\limits_{0}^{\frac{3\pi }{4}}{\sin \left( x+\dfrac{\pi }{4} \right)dx-\sqrt{2}}\int\limits_ {\frac{3\pi }{4}}^{\pi }{\sin \left( x+\dfrac{\pi }{4} \right)dx} \\ & =\sqrt{2}\cos \left( x+\dfrac{\pi }{4} \right)\left| _{\frac{3\pi }{4}}^{\begin{smallmatrix} \pi\\\\\\ \\ \end{smallmatrix}} \right.-\sqrt{2}\cos \left( x+\dfrac{\pi }{4} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{3\pi }{4}} \right. \\ & =\sqrt{2}\left( -\dfrac{\sqrt{2}}{2}+1+1+\dfrac{\sqrt{2}}{2} \right) \\ & =2\sqrt{2} \\ \end{aligned}$