K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

NV
22 tháng 2 2021

\(2I=\int\limits^4_0\left(e^x\sqrt{2x+1}+\dfrac{e^x}{\sqrt{2x+1}}\right)dx=\int\limits^4_0e^x\sqrt{2x+1}dx+\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=I_1+I_2\)

Xét \(I_1=\int\limits^4_0e^x\sqrt{2x+1}dx\)

Đặt \(\left\{{}\begin{matrix}u=\sqrt{2x+1}\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{\sqrt{2x+1}}dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I_1=e^x.\sqrt{2x+1}|^4_0-\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=3e^4-1-I_2\)

Do đó:

\(2I=3e^4-1-I_2+I_2=3e^4-1\)

\(\Rightarrow I=\dfrac{3}{2}e^4-\dfrac{1}{2}\Rightarrow a=\dfrac{3}{2};b=4;c=-\dfrac{1}{2}\)

10 tháng 3 2021

Thanks bạn nhiều

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  
\(x^2-1\) 00 

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

17 tháng 1 2021

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

NV
11 tháng 3 2022

a.

Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)

\(=\dfrac{2}{15}\)

 

NV
11 tháng 3 2022

b.

\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)

Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)

\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)

20 tháng 1 2017

lm jup mk di m.n

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)

\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)

\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)

\(=1+\ln 3\)

AH
Akai Haruma
Giáo viên
7 tháng 2 2017

Câu 1)

\(I=\int \ln ^3 xdx\). Đặt \(\left\{\begin{matrix} u=\ln ^3x\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{3\ln ^2x}{x}dx\\ v=x\end{matrix}\right.\)

\(\Rightarrow I=x\ln ^3x-3\int \ln^2xdx\)

Tiếp tục nguyên hàm từng phần cho \(\int \ln ^2xdx\) như trên, ta suy ra:

\(\int\ln ^2xdx=x\ln^2x-2\int \ln x dx\).

Tiếp tục nguyên hàm từng phần cho \(\int \ln xdx\Rightarrow \int \ln xdx=x\ln x-x+c\)

Do đó mà \(I=x\ln ^3x-3(x\ln^2x-2x\ln x+2x)+c\)

\(\Leftrightarrow I=x\ln^3x-3x\ln^2x+6x\ln x-6x+c\)

AH
Akai Haruma
Giáo viên
7 tháng 2 2017

Câu 2)

\(I=\int ^{1}_{0}(x+\sin ^2x)\cos x dx=\int ^{1}_{0}x\cos xdx+\int ^{1}_{0}\sin^2x\cos xdx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c\)

\(\Rightarrow \int ^{1}_{0} x\cos xdx=\sin 1+\cos 1-1\)

Còn \(\int ^{1}_{0}\sin^2x\cos xdx=\int ^{1}_{0}\sin ^2xd(\sin x)=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\sin ^3x}{3}=\frac{\sin^31}{3}\)

\(\Rightarrow I=-1+\sin 1+\cos 1+\frac{\sin ^3 1}{3}\approx 0,0173\)

NV
23 tháng 1 2019

a/ \(I=\int\limits^1_0\dfrac{1}{\left(x^2+3\right)\left(x^2+1\right)}dx=\dfrac{1}{2}\int\limits^1_0\left(\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}\right)dx\)

\(=\dfrac{1}{2}\left(arctanx-\dfrac{1}{\sqrt{3}}arctan\dfrac{x}{\sqrt{3}}\right)|^1_0=\dfrac{\pi}{8}-\dfrac{\pi\sqrt{3}}{36}\)

b/ \(I=\int\dfrac{x^2-1}{x^4+1}dx=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}dx\)

Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left(1-\dfrac{1}{x^2}\right)dx=dt\) ; \(x^2+\dfrac{1}{x^2}=t^2-2\)

\(\Rightarrow I=\int\dfrac{dt}{t^2-2}=\int\dfrac{dt}{\left(t-\sqrt{2}\right)\left(t+\sqrt{2}\right)}=\dfrac{1}{2\sqrt{2}}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)dt\)

\(\Rightarrow I=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C\)

c/ \(I=\int\dfrac{dx}{x\left(x^3+1\right)}=\int\dfrac{x^2dx}{x^3\left(x^3+1\right)}\)

Đặt \(x^3+1=t\Rightarrow3x^2dx=dt\)

\(\Rightarrow I=\dfrac{1}{3}\int\dfrac{dt}{\left(t-1\right)t}=\dfrac{1}{3}\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\dfrac{1}{3}ln\left|\dfrac{t-1}{t}\right|+C\)

\(\Rightarrow I=\dfrac{1}{3}ln\left|\dfrac{x^3}{x^3+1}\right|+C\)

d/ \(I=\int\limits^1_0\dfrac{xdx}{x^4+x^2+1}\)

Đặt \(x^2=t\Rightarrow2xdx=dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)

\(I=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{t^2+t+1}=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{\left(t+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{2}{3}\int\limits^1_0\dfrac{dt}{\dfrac{4}{3}\left(t+\dfrac{1}{2}\right)^2+1}\)

Đặt \(t+\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}tanu\Rightarrow dt=\dfrac{\sqrt{3}}{2}.\dfrac{du}{cos^2u}\); \(\left\{{}\begin{matrix}t=0\Rightarrow u=\dfrac{\pi}{6}\\t=1\Rightarrow u=\dfrac{\pi}{3}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{du}{cos^2u\left(tan^2u+1\right)}=\dfrac{\sqrt{3}}{3}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}du=\dfrac{\pi\sqrt{3}}{18}\)

22 tháng 1 2019

giup minh voi