$I=\int\limits {\dfrac{2x+3}{x^2-4x+3}} \, dx$

$\begin{array}{l} \quad \displaystyle\int\dfrac{2x+3}{x^2 - 4x+3}dx\\ = \displaystyle\int\left(\dfrac{2x-4}{x^2 - 4x +3} +\dfrac{7}{x^2 - 4x+3}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 - 4x+3}dx + 7\displaystyle\int\dfrac{1}{(x-1)(x-3)}dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 - 4x+3}dx + 7\displaystyle\int\left(\dfrac{1}{2(x-3)} - \dfrac{1}{2(x-1)}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 - 4x+3}dx + \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-3}dx - \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-1}dx\\ = \displaystyle\int\dfrac{d(x^2 - 4x +3)}{x^2 - 4x +3} +\dfrac{7}{2}\displaystyle\int\dfrac{d(x-3)}{x-3} - \dfrac{7}{2}\displaystyle\int\dfrac{d(x-1)}{x-1}\\ = \ln|x^2 - 4x +3| + \dfrac{7}{2}\ln|x-3| - \dfrac{7}{2}\ln|x-1| + C\\ \end{array}$

Đáp án:

 `I=∫(2x+3)/(x^2-4x+3)dx``=-5/2ln| x-1|+9/2ln| x-3|+C `

Giải thích các bước giải:

`I=∫(2x+3)/(x^2-4x+3)dx`

`=∫(2x+3)/[(x-1)(x-3)]dx`

Xét `(2x+3)/[(x-1)(x-3)]=A/(x-1)+B/(x-3)`

`⇒2x+3=A(x-3)+B(x-1)`

`⇒2x+3=Ax-3A+Bx-B`

`⇒2x+3=(A+B)x-3A-B`

`⇒` ``$\begin{cases}A+B=2\\-3A-B=3\\\end{cases}$ `⇒ `$\begin{cases}A=\frac{-5}{2}\\B=\frac{9}{2}\\\end{cases}$ 

Suy ra: `I=∫((-5/2)/(x-1)+(9/2)/(x-3))dx`

`=-5/2ln| x-1|+9/2ln| x-3|+C `