Giải bài 3.5 trang 164 - SBT Đại số và Giải tích lớp 12
Áp dụng phương pháp tính nguyên hàm từng phần, hãy tính:
a) \(\int{\left( 1-2x \right){{e}^{x}}dx}\)
b) \(\int{x{{e}^{-x}}dx}\)
c) \(\int{x\ln \left( 1-x \right)dx}\)
d) \(\int{x{{\sin }^{2}}xdx}\)
a) \(I=\int{\left( 1-2x \right){{e}^{x}}dx}\)
Đặt \(\left\{ \begin{aligned} & u=1-2x \\ & dv={{e}^{x}}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=-2dx \\ & v={{e}^{x}} \\ \end{aligned} \right.\)
Ta có:
\(\begin{align} & I=\left( 1-2x \right){{e}^{x}}+\int{2{{e}^{x}}dx+C} \\ & \,\,\,={{e}^{x}}-2x{{e}^{x}}+2{{e}^{x}}+C \\ & \,\,\,=\left( 3-2x \right){{e}^{x}}+C \\ \end{align}\)
b) \(J=\int{x{{e}^{-x}}dx}\)
Đặt \(\left\{ \begin{aligned} & u=x \\ & dv={{e}^{-x}}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-{{e}^{-x}} \\ \end{aligned} \right.\)
Ta có:
\(\begin{align} & I=-x{{e}^{-x}}+\int{{{e}^{-x}}dx+C} \\ & \,\,\,=-x{{e}^{-x}}-{{e}^{-x}}+C \\ & \,\,\,=-\left( 1+x \right){{e}^{-x}}+C \\ \end{align}\)
c) \(G=\int{x\ln \left( 1-x \right)dx}\)
Đặt \(\left\{ \begin{aligned} & u=ln (1-x) \\ & dv=xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\dfrac{1}{x-1}dx \\ & v=\dfrac{x^2}{2} \\ \end{aligned} \right.\)
Ta có:
\(\begin{align} & G=\frac{{{x}^{2}}}{2}\ln (1-x)-\frac{1}{2}\int{\frac{{{x}^{2}}}{x-1}dx} \\ & \,\,\,\,\,\,=\frac{{{x}^{2}}}{2}\ln (1-x)-\frac{1}{2}\int{\left( x+1+\frac{1}{x-1} \right)dx} \\ & \,\,\,\,\,\,=\frac{{{x}^{2}}}{2}\ln (1-x)-\frac{1}{2}\left[ \frac{{{x}^{2}}}{2}+x+\ln \left( 1-x \right) \right]+C \\ & \,\,\,\,\,\,=\frac{{{x}^{2}}}{2}\ln (1-x)-\frac{1}{2}\ln \left( 1-x \right)-\frac{1}{4}{{x}^{2}}-\frac{1}{2}x+C \\ \end{align} \)
d)Ta có:
\(\begin{align} &H=\int{x{{\sin }^{2}}xdx}\\ & \,\,\,\,\,= \int{x.\dfrac{1-cos 2x}{2}dx}\\ & \,\,\,\,\,=\dfrac{x^2}{4}-\dfrac{1}{2} \int x \cos 2x dx\\ &\,\,\,\,\,=\dfrac{x^2}{4}-\dfrac{1}{2} I \\ \end{align} \)
Đặt \(\left\{ \begin{aligned} & u=x \\ & dv=\cos 2xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=\frac{1}{2}\sin 2x \\ \end{aligned} \right. \)
Suy ra:
\(\begin{align} & I=\dfrac{1}{2}x\sin 2x-\dfrac{1}{2}\int{\sin 2xdx} \\ & \,\,\,\,=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+C \\ \end{align} \)
Vậy \(H=\dfrac{x^2}{4}-\dfrac{1}{2}\left(\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x\right)+C\)
Phương pháp tính nguyên hàm từng phần:
Nếu hai hàm số \(u=u(x)\) và \(v=v(x)\) có đạo hàm liên tục trên K thì
\(\int{u\left( x \right)v'\left( x \right)dx=u\left( x \right)v\left( x \right)-\int{u'\left( x \right)v\left( x \right)dx}}\)Hay \(\int{udv=uv-\int{vdu}}\)