Tìm nguyên làm các hàm số hữu tỉ sau :
a)
\(\int\frac{3x^2+3x+12}{\left(x-1\right)\left(x+2\right)}dx\)
b) \(\int\frac{x^2+2x+6}{\left(x-1\right)\left(x-2\right)\left(x-4\right)}dx\)
Tìm nguyên làm các hàm số hữu tỉ sau :
a)
\(\int\frac{3x^2+3x+12}{\left(x-1\right)\left(x+2\right)}dx\)
b) \(\int\frac{x^2+2x+6}{\left(x-1\right)\left(x-2\right)\left(x-4\right)}dx\)
a) \(f\left(x\right)=\frac{3x^2+3x+12}{\left(x-1\right)\left(x+2\right)x}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x}=\frac{Ax\left(x+2\right)+Bx\left(x-1\right)+C\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)x}\)
Bằng cách thay các nghiệm thực của mẫu số vào hai tử số, ta có hệ :
\(\begin{cases}x=1\rightarrow18=3A\Leftrightarrow A=6\\x=-2\rightarrow18=6B\Leftrightarrow B=3\\x=0\rightarrow12=-2C\Leftrightarrow=-6\end{cases}\) \(\Rightarrow f\left(x\right)=\frac{6}{x-1}+\frac{3}{x+2}-\frac{6}{x}\)
Vậy : \(\int\frac{3x^2+3x+12}{\left(x-1\right)\left(x+2\right)x}dx=\int\left(\frac{6}{x-1}+\frac{3}{x+2}-\frac{6}{x}\right)dx=6\ln\left|x-1\right|+3\ln\left|x+2\right|-6\ln\left|x\right|+C\)
b) \(f\left(x\right)=\frac{x^2+2x+6}{\left(x-1\right)\left(x-2\right)\left(x-4\right)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-4}\)
\(=\frac{A\left(x-2\right)\left(x-4\right)+B\left(x-1\right)\left(x-4\right)+C\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-4\right)}\)
Bằng cách thay các nghiệm của mẫu số vào hai tử số ta có hệ :
\(\begin{cases}x=1\rightarrow9A=3\Leftrightarrow x=3\\x=2\rightarrow14=-2B\Leftrightarrow x=-7\\x=4\rightarrow30=6C\Leftrightarrow C=5\end{cases}\)
\(\Rightarrow f\left(x\right)=\frac{3}{x-1}-\frac{7}{x-2}+\frac{5}{x-4}\)
Vậy :
\(\int\frac{x^2+2x+6}{\left(x-1\right)\left(x-2\right)\left(x-4\right)}dx=\)\(\int\left(\frac{3}{x-1}+\frac{7}{x-2}+\frac{5}{x-4}\right)dx\)=\(3\ln\left|x-1\right|-7\ln\left|x-2\right|+5\ln\left|x-4\right|+C\)
Tìm các nguyên hàm sau:
a) \(I_1=\int\frac{\left(x^2+3\right)dx}{\sqrt{\left(2x-5\right)^3}}\)
b)\(I_2=\int\frac{dx}{\left(3x-1\right)\ln\left(3x-1\right)}\)
c) \(I_3=\int\frac{\left(x^2+1\right)dx}{\sqrt{x^6-7x^4+x^2}}\)
a) Đặt \(\sqrt{2x-5}=t\) khi đó \(x=\frac{t^2+5}{2}\) , \(dx=tdt\)
Do vậy \(I_1=\int\frac{\frac{1}{4}\left(t^2+5\right)^2+3}{t^3}dt=\frac{1}{4}\int\frac{\left(t^4+10t^2+37\right)t}{t^3}dt\)
\(=\frac{1}{4}\int\left(t^2+10+\frac{37}{t^2}\right)dt=\frac{1}{4}\left(\frac{t^3}{3}+10t-\frac{37}{t}\right)+C\)
Trở về biến x, thu được :
\(I_1=\frac{1}{12}\sqrt{\left(2x-5\right)^3}+\frac{5}{2}\sqrt{2x-5}-\frac{37}{4\sqrt{2x-5}}+C\)
b) \(I_2=\frac{1}{3}\int\frac{d\left(\ln\left(3x-1\right)\right)}{\ln\left(3x-1\right)}=\frac{1}{3}\ln\left|\ln\left(3x-1\right)\right|+C\)
c) \(I_3=\int\frac{1+\frac{1}{x^2}}{\sqrt{x^2-7+\frac{1}{x^2}}}dx=\int\frac{d\left(x-\frac{1}{x}\right)}{\sqrt{\left(x-\frac{1}{2}\right)^2-5}}\)
Đặt \(x-\frac{1}{x}=t\)
\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)
\(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)
1) Tìm nguyên hàm: \(\int\dfrac{dx}{\left(x-1\right)^3\sqrt{x^2+3x+1}}\)
2) Tính tích phân sau: \(\int_0^1\left\{\dfrac{1}{x}\right\}\left(\dfrac{x}{1-x}\right)dx\) (kí hiệu \(\left\{a\right\}\) là phần lẻ của số thực \(a\))
Tính nguyên hàm I = \(\int\left(x^2+2x\right)ln\left(3x+1\right)dx\)
Cho biết \(30x^{\dfrac{1}{3}}y^{\dfrac{2}{3}}=360\). Tìm \(\dfrac{dy}{dx}\left(27,8\right)\)
\int \frac{x^3+2x^2+3x-2}{\left(x^2+2x+2\right)^2}dx
\(\int\left(x^2+3x-5\right)\left(2x-3\right)^{10}dx\)
Lời giải:
Đặt \(u=x^2+3x-5; dv=(2x-3)^{10}dx\)
\(\Rightarrow du=(2x+3)dx; v=\int (2x-3)^{10}dx=\frac{1}{2}\int (2x-3)^{10}d(2x-3)=\frac{1}{22}(2x-3)^{11}\)
Do đó:
\(\int (x^2+3x-5)(2x-3)^{10}dx=\frac{1}{22}.(x^2+3x-5)(2x-3)^{11}-\frac{1}{22}\int (2x-3)^{11}(2x+3)dx\)
\(=\frac{1}{22}.(x^2+3x-5)(2x-3)^{11}-\frac{1}{22}[\int (2x-3)^{12}dx+6\int (2x-3)^{11}dx]\)
\(=\frac{1}{22}.(x^2+3x-5)(2x-3)^{11}-\frac{1}{22}[\frac{1}{2}\int (2x-3)^{12}d(2x-3)+3\int (2x-3)^{11}d(2x-3)]\)
\(=\frac{1}{22}(x^2+3x-5)(2x-3)^{11}-\frac{1}{44}.\frac{1}{13}(2x-3)^{13}-\frac{3}{22}.\frac{1}{12}(2x-3)^{12}+C\)
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/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)
cho \(\int f\left(4x\right)dx\) = x2+3x+C. Mệnh đề nào sau đây đúng?
A. \(\int f\left(x+2\right)dx\) =x2+7x+C
B.\(\int f\left(x+2\right)dx\) =\(\frac{x^2}{2}\)+4x+C
C.\(\int f\left(x+2\right)dx\)=\(\frac{x^2}{4}\)+2x+C
D.\(\int f\left(x+2\right)dx\)=\(\frac{x^2}{4}\)+4x+C
Giúp mình bài này với, cám ơn mọi người nhiều
\(\int f\left(4x\right)dx=\frac{1}{4}\int f\left(4x\right)d\left(4x\right)=\frac{1}{16}\left(4x\right)^2+\frac{3}{4}\left(4x\right)+C\)
\(\Rightarrow\int f\left(4x\right)d\left(4x\right)=\frac{1}{4}\left(4x\right)^2+3.\left(4x\right)+C\)
\(\Rightarrow\int f\left(x+2\right)dx=\int f\left(x+2\right)d\left(x+2\right)=\frac{1}{4}\left(x+2\right)^2+3\left(x+2\right)+C\)
\(=\frac{1}{4}x^2+4x+C\)
Cho hàm số f(x) có đạo hàm liên tục trên R. Biết f(3) = 1 và \(\int\limits^1_0xf\left(3x\right)dx=1\) , khi đó \(\int_0^3x^2f'\left(x\right)dx\)
Xét \(I=\int\limits^1_0x.f\left(3x\right)dx\)
Đặt \(3x=u\Rightarrow dx=\dfrac{1}{3}du\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=1\Rightarrow u=3\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{9}\int\limits^3_0u.f\left(u\right)du=\dfrac{1}{9}\int\limits^3_0x.f\left(x\right)dx=1\)
\(\Rightarrow J=\int\limits^3_0x.f\left(x\right)dx=9\)
Xét J, đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{x^2}{2}\end{matrix}\right.\)
\(\Rightarrow J=\dfrac{x^2}{2}.f\left(x\right)|^3_0-\dfrac{1}{2}\int\limits^3_0x^2.f'\left(x\right)dx=\dfrac{9}{2}-\dfrac{1}{2}\int\limits^3_0x^2.f'\left(x\right)dx\)
\(\Rightarrow\int\limits^3_0x^2.f'\left(x\right)dx=9-2J=-9\)