\(\int\limits^{\sqrt{5}}_{\sqrt{2}}\left(2x+\sqrt{x^2-1}\right)dx\)
Những câu hỏi liên quan
Tính tích phân bằng định nghĩa và các tính chất:
1. \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx\)
2. \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx\)
3. \(\int\limits^2_1\frac{2x^3-4x+5}{x}dx\)
4. \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx\)
1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)
2/ \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx=\int\limits^2_1\left(x\sqrt{x}+1\right)dx=\int\limits^2_1\left(x^{\frac{3}{2}}+1\right)dx\)
\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)
3/
\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)
4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)
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Tính các tích phân sau :
a) intlimits^1_0left(y^3+3y^2-2right)dy
b) intlimits^4_1left(t+dfrac{1}{sqrt{t}}-dfrac{1}{t^2}right)dt
c) intlimits^{dfrac{pi}{2}}_0left(2cos x-sin2xright)dx
d) intlimits^1_0left(3^s-2^sright)^2ds
e) intlimits^{dfrac{pi}{3}}_0cos3xdx+intlimits^{dfrac{3pi}{2}}_0cos3xdx+intlimits^{dfrac{5pi}{2}}_{dfrac{3pi}{2}}cos3xdx
g) intlimits^3_0left|x^2-x-2right|dx
h) intlimits^{dfrac{5pi}{4}}_{pi}dfrac{sin x-cos x}{sqrt{1+sin2x}}dx
i) intlimits^4_0dfrac{4x-1}{sqrt{2x+1}+2}dx
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^1_0\left(y^3+3y^2-2\right)dy\)
b) \(\int\limits^4_1\left(t+\dfrac{1}{\sqrt{t}}-\dfrac{1}{t^2}\right)dt\)
c) \(\int\limits^{\dfrac{\pi}{2}}_0\left(2\cos x-\sin2x\right)dx\)
d) \(\int\limits^1_0\left(3^s-2^s\right)^2ds\)
e) \(\int\limits^{\dfrac{\pi}{3}}_0\cos3xdx+\int\limits^{\dfrac{3\pi}{2}}_0\cos3xdx+\int\limits^{\dfrac{5\pi}{2}}_{\dfrac{3\pi}{2}}\cos3xdx\)
g) \(\int\limits^3_0\left|x^2-x-2\right|dx\)
h) \(\int\limits^{\dfrac{5\pi}{4}}_{\pi}\dfrac{\sin x-\cos x}{\sqrt{1+\sin2x}}dx\)
i) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+2}dx\)
Câu nào mình biết thì mình làm nha.
1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
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Tính :
a) intlimits^2_{-1}left(5x^2-x+e^{0,5x}right)dx
b) intlimits^2_{0,5}left(2sqrt{x}+dfrac{3}{x^2}+cos xright)dx
c) intlimits^2_1dfrac{dx}{sqrt{2x+3}} (đặt tsqrt{2x+3} )
d) intlimits^2_1sqrt[3]{3x^3+4}x^2dx (đặt tsqrt[3]{3x^3+4} )
e) intlimits^2_{-2}left(x-2right)left|xright|dx
g) intlimits^0_1xcos xdx
h) intlimits^{dfrac{pi}{2}}_{dfrac{pi}{6}}dfrac{1+sin2x+cos2x}{sin x+cos x}dx
i) intlimits^{dfrac{pi}{2}}_0e^xsin xdx
k) intlimits^e_1x^2ln^2xdx
Đọc tiếp
Tính :
a) \(\int\limits^2_{-1}\left(5x^2-x+e^{0,5x}\right)dx\)
b) \(\int\limits^2_{0,5}\left(2\sqrt{x}+\dfrac{3}{x^2}+\cos x\right)dx\)
c) \(\int\limits^2_1\dfrac{dx}{\sqrt{2x+3}}\) (đặt \(t=\sqrt{2x+3}\) )
d) \(\int\limits^2_1\sqrt[3]{3x^3+4}x^2dx\) (đặt \(t=\sqrt[3]{3x^3+4}\) )
e) \(\int\limits^2_{-2}\left(x-2\right)\left|x\right|dx\)
g) \(\int\limits^0_1x\cos xdx\)
h) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{1+\sin2x+\cos2x}{\sin x+\cos x}dx\)
i) \(\int\limits^{\dfrac{\pi}{2}}_0e^x\sin xdx\)
k) \(\int\limits^e_1x^2\ln^2xdx\)
Tính các tích phân sau
1.Iintlimits^{frac{Pi}{4}}_0 (x+1)sin2xdx
2.Iintlimits^2_1frac{x^2+3x+1}{x^2+x}dx
3.Iintlimits^2_1frac{x^2-1}{x^2}lnxdx
4. Iintlimits^1_0xsqrt{2-x^2}dx
5.Iintlimits^1_0frac{left(x+1right)^2}{x^2+1}dx
6. Iintlimits^5_1frac{dx}{1+sqrt{2x-1}}
7. Iintlimits^3_1frac{1+lnleft(x+1right)}{x^2}dx
8.Iintlimits^1_0frac{x^3}{x^4+3x^2+2}dx
9. Iintlimits^{frac{Pi}{4}}_0xleft(1+sin2xright)dx
10. Iintlimits^3_0frac{x}{sqrt{x+1}}dx
Đọc tiếp
Tính các tích phân sau
1.I=\(\int\limits^{\frac{\Pi}{4}}_0\) (x+1)sin2xdx
2.I=\(\int\limits^2_1\frac{x^2+3x+1}{x^2+x}dx\)
3.I=\(\int\limits^2_1\frac{x^2-1}{x^2}lnxdx\)
4. I=\(\int\limits^1_0x\sqrt{2-x^2}dx\)
5.I=\(\int\limits^1_0\frac{\left(x+1\right)^2}{x^2+1}dx\)
6. I=\(\int\limits^5_1\frac{dx}{1+\sqrt{2x-1}}\)
7. I=\(\int\limits^3_1\frac{1+ln\left(x+1\right)}{x^2}dx\)
8.I=\(\int\limits^1_0\frac{x^3}{x^4+3x^2+2}dx\)
9. I=\(\int\limits^{\frac{\Pi}{4}}_0x\left(1+sin2x\right)dx\)
10. I=\(\int\limits^3_0\frac{x}{\sqrt{x+1}}dx\)
https://i.imgur.com/Pe6vPSJ.jpg
Tính các tích phân sau :
a) intlimits^4_{-2}left(dfrac{x-2}{x+3}right)^2dx (đặt tx+3 )
b) intlimits^6_{-4}left|x+3right|-left|x-4right|dx
c) intlimits^2_{-3}dfrac{dx}{sqrt{x+7}+3} (đặt tsqrt{x+7} hoặc tsqrt{x+7}+3 )
d) intlimits^{dfrac{pi}{2}}_0dfrac{cos x}{1+4sin x}dx
e) intlimits^2_1dfrac{x^9}{x^{10}+4x^5+4}dx (đặt tx^5 )
g) intlimits^3_0left(x+2right)e^{2x}dx
h) intlimits^5_2dfrac{sqrt{4+x}}{x}dx (đặt tsqrt{4+x} )
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^4_{-2}\left(\dfrac{x-2}{x+3}\right)^2dx\) (đặt \(t=x+3\) )
b) \(\int\limits^6_{-4}\left|x+3\right|-\left|x-4\right|dx\)
c) \(\int\limits^2_{-3}\dfrac{dx}{\sqrt{x+7}+3}\) (đặt \(t=\sqrt{x+7}\) hoặc \(t=\sqrt{x+7}+3\) )
d) \(\int\limits^{\dfrac{\pi}{2}}_0\dfrac{\cos x}{1+4\sin x}dx\)
e) \(\int\limits^2_1\dfrac{x^9}{x^{10}+4x^5+4}dx\) (đặt \(t=x^5\) )
g) \(\int\limits^3_0\left(x+2\right)e^{2x}dx\)
h) \(\int\limits^5_2\dfrac{\sqrt{4+x}}{x}dx\) (đặt \(t=\sqrt{4+x}\) )
\(\int\limits^1_0\left(x-1\right)^3\sqrt{2x-x^2}dx\)
chứng minh:
\(\int\limits^1_0\dfrac{ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\dfrac{3}{4}\int\limits\dfrac{ln\left(1+x\right)}{x}^1_0dx\)
1/ Iintlimits^1_0frac{dx}{sqrt{3+2x-x^2}}2/Jintlimits^1_0xlnleft(2x+1right)dx3/Kintlimits^3_2lnleft(x^3-3x+2right)dx4/Iintlimits^{frac{pi}{6}}_0frac{tan^4xdx}{cos2x}5/Jintlimits^3_1frac{3+lnx}{left(x+1right)^2}dx6/Kintlimits^1_0frac{left(2+xe^xright)}{x^2+2x+1}dx
Đọc tiếp
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 \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
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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:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
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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: \(A=\int ^{3}_{1}\frac{3d(x+1)}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-3}{x+1}=\frac{3}{4}\)
\(B=\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-\ln x}{x+1}+\int ^{3}_{1}\frac{dx}{x(x+1)}=\frac{-\ln 3}{4}+\left.\begin{matrix} 3\\ 1\end{matrix}\right|(\ln |x|-\ln|x+1|)\)
\(B=\frac{-\ln 3}{4}+(\ln 3-\ln 4)+\ln 2=\frac{3}{4}\ln 3-\ln 2\)
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Xem thêm câu trả lời
\(\int\limits^1_{\frac{\sqrt{3}}{3}}\frac{\sqrt{\left(1+x^2\right)^5}}{x^8}dx\)
\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)
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