1/ I=\(\int\limits^1_0\)\(\frac{dx}{\sqrt{3+2x-x^2}}\)
2/J=\(\int\limits^1_0\)\(xln\left(2x+1\right)dx\)
3/K=\(\int\limits^3_2ln\left(x^3-3x+2\right)dx\)
4/I=\(\int\limits^{\frac{\pi}{6}}_0\)\(\frac{tan^4xdx}{cos2x}\)
5/J=\(\int\limits^3_1\)\(\frac{3+lnx}{\left(x+1\right)^2}dx\)
6/K=\(\int\limits^1_0\)\(\frac{\left(2+xe^x\right)}{x^2+2x+1}dx\)
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
Câu 5)
\(J=\underbrace{\int ^{3}_{1}\frac{3dx}{(x+1)^2}}_{A}+\underbrace{\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}}_{B}\)
Ta có: \(A=\int ^{3}_{1}\frac{3d(x+1)}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-3}{x+1}=\frac{3}{4}\)
\(B=\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-\ln x}{x+1}+\int ^{3}_{1}\frac{dx}{x(x+1)}=\frac{-\ln 3}{4}+\left.\begin{matrix} 3\\ 1\end{matrix}\right|(\ln |x|-\ln|x+1|)\)
\(B=\frac{-\ln 3}{4}+(\ln 3-\ln 4)+\ln 2=\frac{3}{4}\ln 3-\ln 2\)
Câu 6)
\(K=\underbrace{\int ^{2}_{0}\frac{2dx}{(x+1)^2}}_{A}+\underbrace{\int ^{1}_{0}\frac{xe^x}{(x+1)^2}dx}_{B}\)
Ta có \(A=\int ^{2}_{0}\frac{2d(x+1)}{(x+1)^2}=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{-2}{x+1}=1\)
Với $B$ :
\(\left\{\begin{matrix} u=xe^x\\ dv=\frac{dx}{(x+1)^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^{x}(x+1)dx\\ v=\frac{-1}{x+1}\end{matrix}\right.\)
\(B=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{-xe^x}{x+1}+\int ^{1}_{0}\frac{e^x(x+1)}{x+1}dx=\frac{-e}{2}+\left.\begin{matrix} 1\\ 0\end{matrix}\right|e^x=\frac{e}{2}-1\)
\(\Rightarrow K=\frac{e}{2}\)
Câu 4)
Đặt \(\tan x=t\Rightarrow \cos^2x=\frac{1}{t^2+1},\sin ^2x=\frac{t^2}{t^2+1}\)
\(I=\int ^{\frac{\sqrt{3}}{3}}_{0}\frac{t^4dt}{(t^2+1)\left [ \frac{1}{t^2+1}-\frac{t^2}{t+1} \right ]}=\int ^{\frac{\sqrt{3}}{3}}_{0}\frac{t^4dt}{1-t^2}\)
\(=\int^{\frac{\sqrt{3}}{3}}_{0}(-1-t^2+\frac{1}{(1-t)(1+t)})dt\)
Đến bước này thì rất dễ rồi.
\(I=\left.\begin{matrix} \frac{\sqrt{3}}{3}\\ 0\end{matrix}\right|\frac{-2t(t^2+3)-3\ln|1-t|+3\ln|t+1|}{6}\approx 0,0169786\)
