Toán $I=\int\limits {\dfrac{2x+3}{x^2-4x+3}} \, dx$ 23/11/2021 By Skylar $I=\int\limits {\dfrac{2x+3}{x^2-4x+3}} \, dx$
$\begin{array}{l} \quad \displaystyle\int\dfrac{2x+3}{x^2 – 4x+3}dx\\ = \displaystyle\int\left(\dfrac{2x-4}{x^2 – 4x +3} +\dfrac{7}{x^2 – 4x+3}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + 7\displaystyle\int\dfrac{1}{(x-1)(x-3)}dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + 7\displaystyle\int\left(\dfrac{1}{2(x-3)} – \dfrac{1}{2(x-1)}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-3}dx – \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-1}dx\\ = \displaystyle\int\dfrac{d(x^2 – 4x +3)}{x^2 – 4x +3} +\dfrac{7}{2}\displaystyle\int\dfrac{d(x-3)}{x-3} – \dfrac{7}{2}\displaystyle\int\dfrac{d(x-1)}{x-1}\\ = \ln|x^2 – 4x +3| + \dfrac{7}{2}\ln|x-3| – \dfrac{7}{2}\ln|x-1| + C\\ \end{array}$ Trả lời
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 ` Trả lời
$\begin{array}{l} \quad \displaystyle\int\dfrac{2x+3}{x^2 – 4x+3}dx\\ = \displaystyle\int\left(\dfrac{2x-4}{x^2 – 4x +3} +\dfrac{7}{x^2 – 4x+3}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + 7\displaystyle\int\dfrac{1}{(x-1)(x-3)}dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + 7\displaystyle\int\left(\dfrac{1}{2(x-3)} – \dfrac{1}{2(x-1)}\right)dx\\ = \displaystyle\int\dfrac{2x-4}{x^2 – 4x+3}dx + \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-3}dx – \dfrac{7}{2}\displaystyle\int\dfrac{1}{x-1}dx\\ = \displaystyle\int\dfrac{d(x^2 – 4x +3)}{x^2 – 4x +3} +\dfrac{7}{2}\displaystyle\int\dfrac{d(x-3)}{x-3} – \dfrac{7}{2}\displaystyle\int\dfrac{d(x-1)}{x-1}\\ = \ln|x^2 – 4x +3| + \dfrac{7}{2}\ln|x-3| – \dfrac{7}{2}\ln|x-1| + C\\ \end{array}$
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 `