Question

Rezolvați integrala :

$\int\limits^4_2 {\frac{x-3}{ (x^{2}-6x+5) ^{2}} } \, dx$

Answer 1

Răspuns:

Ai raspunsul atasat

P.S. se mai poate face intr-un fel . Faci substitutia x-3=u -> u+3=x -> dx =du , capetele vor fi u =-1 si u =1 .( Observatie : pe intervalul [-1,1] functia este impara ..... ceea ce inseamna ca rezultatul va iesi la iveala imediat : 0 )

Explicație pas cu pas:

Answer 2

$\displaystyle \int_{2}^4\dfrac{x-3}{(x^2-6x+5)^2}\, dx =\\ \\ =\dfrac{1}{2}\int_{2}^4\dfrac{2x-6}{(x^2-6x+5)^2}\, dx= \dfrac{1}{2}\int_{2}^4\dfrac{(x^2-6x+5)'}{(x^2-6x+5)^2}\, dx = \\ \\ =\dfrac{1}{2}\int_{2}^4(x^2-6x+5)'\cdot (x^2-6x+5)^{-2}\, dx = \\ \\ = \dfrac{1}{2}\cdot \dfrac{(x^2-6x+5)^{-2+1}}{-2+1}\Bigg|_{2}^{4}=-\dfrac{1}{2(x^2-6x+5)}\Bigg|_{2}^{4}=\\ \\ =-\dfrac{1}{2(4^2-6\cdot 4+5)}+\dfrac{1}{2(2^2-6\cdot 2+5)}=\\ \\=-\dfrac{1}{2\cdot(-3)}-\dfrac{1}{2\cdot 3} = \boxed{0}$