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NT
1
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 3 2022
\(I=\int\limits^e_1x^2.ln^2x.\dfrac{1}{x\left(lnx+1\right)^2}dx\)
Đặt \(\left\{{}\begin{matrix}u=x^2ln^2x\\dv=\dfrac{1}{x\left(lnx+1\right)^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2x.lnx\left(lnx+1\right)\\v=-\dfrac{1}{lnx+1}\end{matrix}\right.\)
\(\Rightarrow I=-\dfrac{x^2ln^2x}{lnx+1}|^e_1+\int\limits^e_12x.lnxdx=-\dfrac{e^2}{2}+I_1\)
Xét \(I_1\), đặt \(\left\{{}\begin{matrix}u=lnx\\dv=2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2\end{matrix}\right.\)
\(\Rightarrow I_1=x^2lnx|^e_1-\int\limits^e_1xdx=...\)
HT
17 tháng 1 2021
Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ
Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)
| x | -2 | -1 | 1 | 2 |
| \(x^2-1\) | 0 | 0 |
\(\left(-2;-1\right):+\)
\(\left(-1;1\right):-\)
\(\left(1;2\right):+\)
\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)
\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)
\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)
Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính
HT
17 tháng 1 2021
2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)
\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)
\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)
22 tháng 3 2016
\(I_1=3\int_1^2x^2dx+\int_1^2\cos xdx+\int_1^2\frac{dx}{x}=x^3\)\(|^2 _1\)+\(\sin x\)\(|^2_1\) +\(\ln\left|x\right|\)\(|^2_1\)
\(=\left(8-1\right)+\left(\sin2-\sin1\right)+\left(\ln2-\ln1\right)\)
\(=7+\sin2-\sin1+\ln2\)
22 tháng 3 2016
b) \(I_2=4\int_1^2\frac{dx}{x}-5\int_1^2x^4dx+2\int_1^2\sqrt{x}dx\)
\(=4\left(\ln2-\ln1\right)-\left(2^5-1^5\right)+\frac{4}{3}\left(2\sqrt{2}-1\sqrt{1}\right)\)
\(=4\ln2+\frac{8\sqrt{2}}{3}-32\frac{1}{3}\)
AH
Akai Haruma
Giáo viên
8 tháng 2 2017
Câu 1)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2\ln x}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)
\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)
Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)
AH
Akai Haruma
Giáo viên
8 tháng 2 2017
Câu 2)
\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)
\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)
QT
1
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2021
\(\dfrac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{\left(x+1\right)^2x-x^2\left(x+1\right)}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{x\left(x+1\right)}\)
\(=\dfrac{\sqrt{x}}{x}-\dfrac{\sqrt{x+1}}{x+1}=x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\)
Do đó:
\(I=\int\limits^2_1\left[x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\right]dx=\left(2\sqrt{x}-2\sqrt{x+1}\right)|^2_1=...\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2022
\(f\left(x\right)-\left(x+1\right)f'\left(x\right)=2x.f^2\left(x\right)\)
\(\Rightarrow\dfrac{f\left(x\right)-\left(x+1\right)f'\left(x\right)}{f^2\left(x\right)}=2x\)
\(\Rightarrow\left[\dfrac{x+1}{f\left(x\right)}\right]'=2x\)
Lấy nguyên hàm 2 vế:
\(\dfrac{x+1}{f\left(x\right)}=\int2xdx=x^2+C\)
Thay \(x=1\Rightarrow\dfrac{2}{f\left(1\right)}=1+C\Rightarrow C=0\)
\(\Rightarrow f\left(x\right)=\dfrac{x+1}{x^2}\Rightarrow\int\limits^2_1\left(\dfrac{1}{x}+\dfrac{1}{x^2}\right)dx=\left(lnx-\dfrac{1}{x}\right)|^2_1=ln2+\dfrac{1}{2}\)
AH
Akai Haruma
Giáo viên
20 tháng 11 2017
Câu a)
\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)
Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)
Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)
Vậy :
\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)
\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)
Câu b)
\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)
\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)
Do đó:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
AH
Akai Haruma
Giáo viên
20 tháng 11 2017
Câu c)
\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)
\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)
\(=\frac{x^2}{2}+c+\ln ^2x\)
\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)
Câu d)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 2 2020
\(I=\int\limits^e_1\frac{\frac{1-lnx}{x^2}}{\left(1+\frac{lnx}{x}\right)^2}dx\)
Đặt \(\frac{lnx}{x}=t\Rightarrow\left(\frac{1-lnx}{x^2}\right)dx=dt\)
\(\Rightarrow I=\int\limits^{\frac{1}{e}}_0\frac{dt}{\left(1+t\right)^2}=-\frac{1}{1+t}|^{\frac{1}{e}}_0=\frac{1}{e+1}\)
\(\Rightarrow a=b=1\Rightarrow a^2+b^2=2\)
\(I=\int\limits^e_1x^2ln^2xdx\) (do \(\left(xlnx\right)^2>0\))
Đặt \(\left\{{}\begin{matrix}u=ln^2x\\dv=x^2dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2lnx}{x}dx\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
\(I=\dfrac{1}{3}x^3ln^2x|^e_1-\dfrac{2}{3}\int\limits^e_1x^2lnxdx=\dfrac{1}{3}e^3-\dfrac{2}{3}I_1\)
Xét \(I_1=\int\limits^e_1x^2lnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=x^2dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
\(I_1=\dfrac{1}{3}x^3lnx|^e_1-\dfrac{1}{3}\int\limits^e_1x^2dx=\dfrac{1}{3}e^3-\dfrac{1}{9}x^3|^e_1=\dfrac{2}{9}e^3+\dfrac{1}{9}\)
\(\Rightarrow I=\dfrac{1}{3}e^3-\dfrac{2}{3}\left(\dfrac{2}{9}e^3+\dfrac{1}{9}\right)=...\)