Toán tìm nguyên hàm của (x^2-5x+8) / ( x-1)*(x^2-4x+5) 25/07/2021 By Arya tìm nguyên hàm của (x^2-5x+8) / ( x-1)*(x^2-4x+5)
Ta có $\int \left( \dfrac{x^2 – 5x + 8}{(x-1)(x^2 – 4x + 5)}\right) dx = \int \left(\dfrac{2-x}{x^2 – 4x + 5} + \dfrac{2}{x-1} \right)$ $= -\dfrac{1}{2} \int \dfrac{2x-4}{x^2 – 4x + 5} dx + \int \dfrac{2}{x-1} dx$ $= – \dfrac{1}{2} \int \dfrac{d(x^2-4x+5)}{x^2-4x+5} dx + 2\ln |x-1| + c$ $= -\dfrac{1}{2} \ln |x^2-4x+5| + 2\ln|x-1| + c$ Trả lời
Ta có
$\int \left( \dfrac{x^2 – 5x + 8}{(x-1)(x^2 – 4x + 5)}\right) dx = \int \left(\dfrac{2-x}{x^2 – 4x + 5} + \dfrac{2}{x-1} \right)$
$= -\dfrac{1}{2} \int \dfrac{2x-4}{x^2 – 4x + 5} dx + \int \dfrac{2}{x-1} dx$
$= – \dfrac{1}{2} \int \dfrac{d(x^2-4x+5)}{x^2-4x+5} dx + 2\ln |x-1| + c$
$= -\dfrac{1}{2} \ln |x^2-4x+5| + 2\ln|x-1| + c$