Ta có :

\(\frac{3x+2}{x^2+2x-3}=\frac{E\left(2x+2\right)+D}{x^2+2x-3}=\frac{2E+D+2E}{x^2+2x-3}\)

Đồng nhất hệ số hai tử sốta có hệ phương trình 

\(\begin{cases}2E=3\\D+2E=2\end{cases}\) \(\Rightarrow\begin{cases}E=\frac{3}{2}\\D=-1\end{cases}\)

\(\Rightarrow\) \(\frac{3x+2}{x^2+2x-3}=\frac{\frac{3}{2}\left(2x+2\right)}{x^2+2x-3}-\frac{1}{x^2+2x-3}\)

Vậy :

\(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\int\frac{d\left(x^2+2x-3\right)}{x^2+2x-3}+\int\frac{1}{x^2+2x-3}dx\)\(=\frac{3}{2}\ln\left|x^2+2x-3\right|+J\left(1\right)\)

Tính :

\(J=\int\frac{1}{x^2+2x-3}dx=\frac{1}{4}\left(\int\frac{1}{x-1}dx-\int\frac{1}{x+3}dx\right)=\frac{1}{4}\ln\left|x-1\right|-\ln\left|x+3\right|=\frac{1}{4}\ln\left|\frac{x-1}{x+3}+C\right|\)

Do đó :  \(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\ln\left|x^2+2x-3\right|+\frac{1}{4}\ln\left|\frac{x-1}{x+3}\right|+C\)

b) Ta có :

\(\frac{2x-3}{x^2+4x+4}=\frac{E\left(2x+4\right)+D}{x^2+4x+4}=\frac{2Ex+D+4E}{x^2+4x+4}\)

Đồng nhất hệ số  hai tử số :

Ta có hệ : \(\Leftrightarrow\)\(\begin{cases}2E=2\\D+4E=-3\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}E=1\\D=-7\end{cases}\)

Suy ra :

\(\frac{2x-3}{x^2+4x+4}=\frac{2x+4}{x^2+4x+4}-\frac{7}{x^2+4x+4}\)

Vậy : \(\int\frac{2x-3}{x^2+4x+4}dx=\int\frac{2x+4}{x^2+4x+4}dx-7\int\frac{1}{\left(x+2\right)^2}dx=\ln\left|x^2+4x+4\right|+\frac{7}{x+2}+C\)

a)

\(\int\frac{2\left(x_{ }+1\right)}{x^2+2x_{ }-3}dx=\int\frac{2x+2}{x^2+2x-3}dx\)

\(=\int\frac{d\left(x^2+2x-3\right)}{x^2+2x-3}=ln\left|x^2+2x-3\right|+C\)

b)\(\int\frac{2\left(x-2\right)dx}{x^2-4x+3}=\int\frac{2x-4dx}{x^2-4x+3}=\int\frac{d\left(x^2-4x+3\right)}{x^2-4x+3}=ln\left|x^2-4x+3\right|+C\)

\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)

a) Mẫu số chứa các biểu thức có nghiệm  thực và không có nghiệm thực.

\(f\left(x\right)=\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A\left(x^2+1\right)+\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}\left(1\right)\)

Tay x=1 vào 2 tử, ta có : 2=2A, vậy A=1

Do đó (1) trở thành : 

\(\frac{1\left(x^2+1\right)+\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(B+1\right)x^2+\left(C-B\right)x+1-C}{\left(x-1\right)\left(x^2+1\right)}\)

Đồng nhất hệ số hai tử số, ta có hệ :

\(\begin{cases}B+1=1\\C-B=2\\1-C=-1\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}B=0\\C=2\\A=1\end{cases}\)\(\Rightarrow\)

\(f\left(x\right)=\frac{1}{x-1}+\frac{2}{x^2+1}\)

Vậy :

\(f\left(x\right)=\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}dx=\int\frac{1}{x-1}dx+2\int\frac{1}{x^2+1}=\ln\left|x+1\right|+2J+C\left(2\right)\)

* Tính \(J=\int\frac{1}{x^2+1}dx.\)

Đặt \(\begin{cases}x=\tan t\rightarrow dx=\left(1+\tan^2t\right)dt\\1+x^2=1+\tan^2t\end{cases}\)

Cho nên :

\(\int\frac{1}{x^2+1}dx=\int\frac{1}{1+\tan^2t}\left(1+\tan^2t\right)dt=\int dt=t;do:x=\tan t\Rightarrow t=arc\tan x\)

Do đó, thay tích phân J vào (2), ta có : 

\(\int\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}dx=\ln\left|x-1\right|+arc\tan x+C\)

b) Ta phân tích 

\(f\left(x\right)=\frac{x^2+1}{\left(x-1\right)^3\left(x+3\right)}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}+\frac{D}{x+3}\)\(=\frac{A\left(x+3\right)+B\left(x-1\right)\left(x+3\right)+C\left(x-1\right)^2\left(x+3\right)+D\left(x-1\right)^3}{\left(x-1\right)^3\left(x+3\right)}\)

Thay x=1 và x=-3 vào hai tử số, ta được :

\(\begin{cases}x=1\rightarrow2=4A\rightarrow A=\frac{1}{2}\\x=-3\rightarrow10=-64D\rightarrow D=-\frac{5}{32}\end{cases}\)

Thay hai giá trị của A và D vào (*) và đồng nhất hệ số hai tử số, ta cso hệ hai phương trình :

\(\begin{cases}0=C+D\Rightarrow C=-D=\frac{5}{32}\\1=3A-3B+3C-D\Rightarrow B=\frac{3}{8}\end{cases}\)

\(\Rightarrow f\left(x\right)=\frac{1}{2\left(x-1\right)^3}+\frac{3}{8\left(x-1\right)^2}+\frac{5}{32\left(x-1\right)}-+\frac{5}{32\left(x+3\right)}\)

Vậy : 

\(\int\frac{x^2+1}{\left(x-1\right)^3\left(x+3\right)}dx=\)\(\left(\frac{1}{2\left(x-1\right)^3}+\frac{3}{8\left(x-1\right)^2}+\frac{5}{32\left(x-1\right)}-+\frac{5}{32\left(x+3\right)}\right)dx\)

\(=-\frac{1}{a\left(x-1\right)^2}-\frac{3}{8\left(x-1\right)}+\frac{5}{32}\ln\left|x-1\right|-\frac{5}{32}\ln\left|x+3\right|+C\)

\(=-\frac{1}{a\left(x-1\right)^2}-\frac{3}{8\left(x-1\right)}+\frac{5}{32}\ln\left|\frac{x-1}{x+3}\right|+C\)

cac cau nghi gi ve cau hoi nay

a) \(\int\frac{1}{x^2-3x+2}dx=\frac{1}{2-1}\int\frac{1}{\left(x-1\right)\left(x-2\right)}dx\)

=\(\int\frac{1}{x-2}dx-\int\frac{1}{x-1}dx=ln\left|x-2\right|-ln\left|x-1\right|=ln\left|\frac{x-2}{x-1}+C\right|\)

b) \(\int\frac{1}{4x^2-3x-1}dx=\frac{1}{4}.\frac{1}{\left(1-\frac{1}{4}\right)}\int\frac{1}{\left(x+\frac{1}{4}\right)\left(x-1\right)}dx\)

=\(\frac{1}{3}.\left[\int\frac{1}{x-1}dx-\int\frac{1}{x+\frac{1}{4}}dx\right]\)

=\(\frac{1}{3}\left[ln\left|x-1\right|-ln\left|x+\frac{1}{4}\right|\right]=\frac{1}{3}ln\left|\frac{x-1}{x+\frac{1}{4}}\right|+C\)

=\(\frac{1}{3}ln\left|\frac{4\left(x-1\right)}{4x+1}+C\right|\)

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 \(x-\frac{1}{x}=t\)

\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)

                           \(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)

Chịu thôi khó quá.

Ta có :\(x^3-2x^2-x+2=x\left(x^2-1\right)-2\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

Ta viết biểu thức dạng \(\frac{x^2-3}{x^3-2x^2-x+2}=\frac{A_1}{x+1}+\frac{A_2}{x-1}+\frac{A_3}{x-2}\)

Từ đó 

\(A_1\left(x-1\right)\left(x-2\right)+A_2\left(x+1\right)\left(x-2\right)+A_3\left(x+1\right)\left(x-1\right)\equiv x^2-3\) (1)

hay là \(\left(A_1+A_2+A_3\right)x^2+\left(-3A_1-A_2\right)x+\left(2A_1-2A_2-A_3\right)\equiv x^2-3\)

Áp dụng phương pháp cân bằng hệ số ta có

\(x^2\)  \(A_1+A_2+A\)

\(x^1\)  \(-3A_1-A\)

\(x^0\)  \(2A_1-2A_2-A\)

\(\Rightarrow A_1=-\frac{1}{3},A_2=1,A_3=\frac{1}{3}\)

a)

\(\frac{1}{x^2-4x+4}dx=\frac{1}{\left(x-2\right)^2}dx=-\frac{1}{x-2}+C\)

b) \(\frac{1}{9x^2-12x+4}dx=\frac{1}{9\left(x-\frac{2}{3}\right)^2}dx=\frac{1}{9}.\frac{1}{\left(x-\frac{2}{3}\right)^2}dx=\frac{1}{9}.\frac{1}{x-\frac{2}{3}}=\frac{1}{9x-6}+C\)