Bài 4 (Trang 113 SGK Toán Giải tích 12):

Sử dụng phương pháp tích phân từng phần, tính các tích phân:

$$\begin{gathered} a) {\int }_0^{\frac{\pi }{2}}(x+1\left)\sin xdx \\ b\right) {\int }_1^e{x}^2\ln xdx \\ c) {\int }_0^1\ln (1+x\left)dx \\ d\right) {\int }_0^1(x^2-2x-1)e^{-x}dx. \end{gathered}$$

Hướng dẫn giải:

a) Đặt $\left\{\begin{array}{l}\mathrm{u}=\mathrm{x}+1\\ \mathrm{dv}=\mathrm{sinxdx}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{du}=\mathrm{dx}\\ \mathrm{v}=-\mathrm{cosx}\end{array}\right.\right.$

${\int }_0^{\frac{\mathrm{\pi }}{2}}(\mathrm{x}+1)\mathrm{sinxdx} =-(\mathrm{x}+1)\mathrm{cosx}{|}_0^{\frac{\mathrm{\pi }}{2}} +{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{cosxdx} =-(\mathrm{x}+1)\mathrm{cosx}{|}_0^{\frac{\mathrm{\pi }}{2}} +\mathrm{sinx}{|}_0^{\frac{\mathrm{\pi }}{2}} = 2$

b) Đặt $\left\{\begin{array}{l}\mathrm{u}=\mathrm{lnx}\\ \mathrm{dv}={\mathrm{x}}^2\mathrm{dx}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{du}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{v}=\frac{{\mathrm{x}}^3}{3}\end{array}\right.\right.$

${\int }_1^{\mathrm{e}}{\mathrm{x}}^2\mathrm{lnxdx} = \frac{{\mathrm{x}}^3}{3}\mathrm{lnx}{|}_1^{\mathrm{e}}-\frac{1}{3}{\int }_1^{\mathrm{e}}{\mathrm{x}}^2\mathrm{dx} = \frac{{\mathrm{x}}^3}{3}\mathrm{lnx}{|}_1^{\mathrm{e}}-\frac{{\mathrm{x}}^3}{9}{|}_1^{\mathrm{e}} = \frac{1}{9}(2{\mathrm{e}}^3+1)$

c) Đặt $\left\{\begin{array}{l}\mathrm{u}=\ln (1+\mathrm{x})\\ \mathrm{dv}=\mathrm{dx}\end{array} \Rightarrow \left\{\begin{array}{l}\mathrm{du} = \frac{\mathrm{dx}}{1+\mathrm{x}}\\ \mathrm{v} = \mathrm{x}\end{array}\right.\right.$

${\int }_0^1\ln (1+\mathrm{x})\mathrm{dx} = \mathrm{xln}(1+\mathrm{x}){|}_0^1-{\int }_0^1\frac{\mathrm{x}}{1+\mathrm{x}}\mathrm{dx} = \mathrm{xln}(1+\mathrm{x}){|}_0^1-{\int }_0^1(1-\frac{\mathrm{x}}{\mathrm{x}+1})\mathrm{dx}$

$= \mathrm{xln}(1+\mathrm{x}){|}_0^1 - {\int }_0^1(1-\frac{1}{\mathrm{x}+1})\mathrm{dx} = \ln 2-(\mathrm{x}-\ln (1+\mathrm{x}){|}_0^1 = 2\ln 2-1$

d) Đặt $\left\{\begin{array}{l}\mathrm{u}={\mathrm{x}}^2-2\mathrm{x}-1\\ \mathrm{dv}={\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{du}=(2\mathrm{x}-2)\mathrm{dx}\\ \mathrm{v}=-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right.\right.$

   $\Rightarrow {\int }_0^1({\mathrm{x}}^2-2\mathrm{x}-1){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx} = -({\mathrm{x}}^2-2\mathrm{x}-1){\mathrm{e}}^{-\mathrm{x}}{|}_0^1+{\int }_0^1(2\mathrm{x}-2){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}$

    Đặt $\left\{\begin{array}{l}\mathrm{u}=2\mathrm{x}-2\\ \mathrm{dv}={\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{du}=2\mathrm{dx}\\ \mathrm{v}=-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right.\right.$

  $\Rightarrow {\int }_0^1(2x-2)e^{-x}dx=-(2x-2)e^{-x}{|}_0^1+2{\int }_{}^{}{e}^{-x}dx=-(2x-2)e^{-x}{|}_0^1-2e^{-x}{|}_0^1$

Vậy ${\int }_0^1({\mathrm{x}}^2-2\mathrm{x}-1){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx} = -({\mathrm{x}}^2-2\mathrm{x}-1){\mathrm{e}}^{-\mathrm{x}}{|}_0^1 -(2\mathrm{x}-2){\mathrm{e}}^{-\mathrm{x}}{|}_0^1 -2{\mathrm{e}}^{-\mathrm{x}}{|}_0^1 =-1$