Lời giải:
a) Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos 2xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{\sin 2x}{2}\end{matrix}\right.\)
\(\Rightarrow \int x\cos 2xdx=\frac{x\sin 2x}{2}-\int \frac{\sin 2x}{2}dx=\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}\)
\(\Rightarrow \int ^{\frac{\pi}{2}}_{0}x\cos 2xdx=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\left ( \frac{x\sin 2x}{2}+\frac{\cos 2x}{4} \right )=\frac{-1}{2}\)
b) Đặt \(\left\{\begin{matrix} u=x\\ dv=e^{-2x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{-e^{-2x}}{2}\end{matrix}\right.\)
\(\Rightarrow \int xe^{-2x}dx=\frac{-xe^{-2x}}{2}+\int \frac{e^{-2x}}{2}dx=\frac{-xe^{-2x}}{2}-\frac{e^{-2x}}{4}\)
\(\Rightarrow \int ^{\ln 2}_{0}xe^{-2x}dx=\left.\begin{matrix} \ln 2\\ 0\end{matrix}\right|\left ( \frac{-xe^{-2x}}{2}-\frac{e^{2x}}{4} \right )=\frac{3}{16}-\frac{\ln 2}{8}\)
c)
\(\int ^{1}_{0}\ln (2x+1)dx=\frac{1}{2}\int ^{1}_{0}\ln (2x+1)d(2x+1)=\frac{1}{2}\int ^{3}_{1}\ln tdt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t\)
Do đó \(\frac{1}{2}\int ^{3}_{1}\ln tdt=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\left(\frac{t\ln t-t}{2}\right)=\frac{3\ln 3}{2}-1\)
d)
Ta có \(\int ^{3}_{2}(\ln (x-1)-\ln (x+1))dx=\int ^{3}_{2}\ln (x-1)d(x-1)-\int ^{3}_{2}\ln (x+1)d(x+1)\)
\(=\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt\)
Theo phần c, ta đã chỉ ra được \(\int \ln tdt=t\ln t-t\), do đó:
\(\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|(t\ln t-t)-\left.\begin{matrix} 4\\ 3\end{matrix}\right|(t\ln t-t)=\ln \left(\frac{27}{64}\right)\)
e)
Xét \(\int (x+1-\frac{1}{x})e^{x+\frac{1}{x}}dx=\int e^{x+\frac{1}{x}}dx+\int \left (x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)
Đặt \(\left\{\begin{matrix} u=e^{x+\frac{1}{x}}\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx\\ v=x\end{matrix}\right.\)
\(\Rightarrow \int e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}-\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)
Do đó \(\int \left(x+1-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}\)
\(\int ^{2}_{\frac{1}{2}}\left(x+1-\frac{x}{x}\right)e^{x+\frac{1}{x}}dx=\left.\begin{matrix} 2\\ \frac{1}{2}\end{matrix}\right|xe^{x+\frac{1}{x}}=\frac{3e^{\frac{5}{2}}}{2}\)
g)
Có \(\int x\cos x\sin^2xdx=\frac{1}{2}\int x\sin 2x\sin xdx=\frac{1}{4}\int x(\cos x-\cos 3x)dx\)
Xét \( \int x\cos xdx\). Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\)
\(\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x\)
Tương tự, \(\int x\cos 3xdx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}\)
Do đó, \(\int ^{\frac{\pi}{2}}_{0}x\cos x\sin^2xdx=\int ^{\frac{\pi}{2}}_{0}\frac{1}{4}x(\cos x-\cos 3x)dx\)
\(=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|(x\sin x+\cos x-\frac{x\sin 3x}{3}-\frac{\cos 3x}{9})=\frac{3\pi-4}{18}\)
h)
\(\left\{\begin{matrix} u=xe^x\\ dv=\frac{1}{(x+1)^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^x(x+1)dx\\ v=\frac{-1}{x+1}\end{matrix}\right.\)
\(\Rightarrow \int \frac{xe^x}{(x+1)^2}dx=\frac{-xe^x}{x+1}+\int e^xdx=\frac{-xe^x}{x+1}+e^x=\frac{e^x}{x+1}\)
Do đó \(\int ^{1}_{0}\frac{xe^x}{(x+1)^2}dx=\frac{e}{2}-1\)
i)
Có \(\int \frac{1+x\ln x}{x}e^xdx=\int \frac{e^x}{x}dx+\int \ln xe^xdx\)
Đặt \(\left\{\begin{matrix} u=\ln x\\ dv=e^xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=e^x\end{matrix}\right.\Rightarrow \int \ln xe^xdx=\ln xe^x-\int \frac{e^x}{x}dx\)
\(\Rightarrow \int ^{e}_{1}\frac{1+x\ln x}{x}e^xdx=\left.\begin{matrix} e\\ 1\end{matrix}\right|\ln xe^x=e^e\)
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