Giải bài 3.9 trang 165 - SBT Đại số và Giải tích lớp 12
Tính các nguyên hàm sau đây:
a) \(\int(x+ln x)x^2dx\)
b) \(\int(x+\sin^2x)\sin x\,dx\)
c) \(\int(x+e^x)e^{2x}\,dx\)
d) \(\int(x+\sin x)\dfrac{dx}{\cos^2x}\)
a) \(I=\int(x+ln x)x^2dx\)
Đặt \(\left\{ \begin{aligned} & u=x+\ln x \\ & dv={{x}^{2}}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=1+\frac{1}{x} \\ & v=\frac{1}{3}{{x}^{3}} \\ \end{aligned} \right. \)
Ta có:
\(\begin{align} & I=\frac{1}{3}{{x}^{3}}\left( x+\ln x \right)-\frac{1}{3}\int{{{x}^{3}}\left( 1+\frac{1}{x} \right)dx} \\ & \,\,\,=\frac{1}{3}{{x}^{4}}+\frac{1}{3}{{x}^{3}}\ln x-\frac{1}{3}\int{\left( {{x}^{3}}+{{x}^{2}} \right)} \\ & \,\,\,=\frac{1}{3}{{x}^{4}}+\frac{1}{3}{{x}^{3}}\ln x-\frac{1}{3}\left( \frac{1}{4}{{x}^{4}}+\frac{1}{3}{{x}^{3}} \right)+C \\ & \,\,\,\,=\frac{1}{4}{{x}^{4}}-\frac{1}{9}{{x}^{3}}+\frac{1}{3}{{x}^{3}}\ln x+C \\ \end{align} \)
b) \(J=\int(x+\sin^2x)\sin x\,dx\)
Đặt \(\left\{ \begin{aligned} & u=x+\sin^2 x\\ & dv=\sin xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=1+2 \sin x \cos x \\ & v=-\cos x \\ \end{aligned} \right. \)
Ta có:
\(\begin{align} & J=-\cos x\left( x+{{\sin }^{2}}x \right)+\int{\cos x\left( 1+2\sin x\cos x \right)dx} \\ & \,\,\,=-x\cos x-\cos x{{\sin }^{2}}x+\int{\cos xdx+2\int{{{\cos }^{2}}x\sin xdx}} \\ & \,\,\,=-x\cos x-\cos x{{\sin }^{2}}x+\sin x-2\int{{{\cos }^{2}}x\,d\left( \cos x \right)} \\ & \,\,\,=-x\cos x-\cos x{{\sin }^{2}}x+\sin x-\frac{2}{3}{{\cos }^{3}}x+C \\ \end{align} \)
c) \(K=\int(x+e^x)e^{2x}\,dx\)
Đặt \(\left\{ \begin{aligned} & u=x+e^x\\ & dv=e^{2x}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=1+e^x \\ & v=\dfrac{1 }{2}e^{2x}\\ \end{aligned} \right. \)
Ta có:
\(\begin{align} & K=\frac{1}{2}\left( x+{{e}^{x}} \right){{e}^{2x}}-\frac{1}{2}\int{{{e}^{2x}}\left( 1+{{e}^{x}} \right)dx} \\ & \,\,\,\,\,\,=\frac{1}{2}\left( x+{{e}^{x}} \right){{e}^{2x}}-\frac{1}{2}\int{\left( {{e}^{2x}}+{{e}^{3x}} \right)dx} \\ & \,\,\,\,\,\,=\frac{1}{2}x{{e}^{2x}}+\frac{1}{2}{{e}^{3x}}-\frac{1}{4}{{e}^{2x}}-\frac{1}{6}{{e}^{3x}}+C \\ & \,\,\,\,\,=\frac{1}{2}x{{e}^{2x}}+\frac{1}{3}{{e}^{3x}}-\frac{1}{4}{{e}^{2x}}+C \\ \end{align} \)
d) \(F=\int(x+\sin x)\dfrac{dx}{\cos^2x}\)
Đặt \(\left\{ \begin{aligned} & u=x+\sin x \\ & dv=\frac{dx}{{{\cos }^{2}}x} \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=1+\cos x \\ & v=\tan x \\ \end{aligned} \right. \)
Ta có:
\(\begin{align} & F=\left( x+\sin x \right)\tan x-\int{\left( 1+\cos x \right)\tan xdx} \\ & \,\,\,\,=\left( x+\sin x \right)\tan x-\int{\left( \frac{\sin x}{\cos x}+\sin x \right)dx} \\ & \,\,\,\,=\left( x+\sin x \right)\tan x+\ln \left| \cos x \right|+\cos x+C \\ \end{align} \)