∫((1−x^2 −8x+1)/(x^4 −x^3 −x+1))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 76468 by kaivan.ahmadi last updated on 27/Dec/19

$\int \frac{1 - x^{2} - 8 x + 1}{x^{4} - x^{3} - x + 1} d x$
Commented by mr W last updated on 27/Dec/19

$w h y d i d y o u w r i t e 1 - x^{2} - 8 x + 1$ $i n s t e a d o f 2 - x^{2} - 8 x ?$
Commented by kaivan.ahmadi last updated on 27/Dec/19

$i d o n t k n o w . i t i s a p r a c t i c e t h a t i s a w i t .$
	Answered by MJS last updated on 27/Dec/19

	$just decompose$ $\int \frac{- x^{2} - 8 x + 2}{x^{4} - x^{3} - x + 1} d x = - \int \frac{x^{2} + 8 x - 2}{{(x - 1)}^{2} (x^{2} + x + 1)} d x =$ $= \int \frac{x + \frac{10}{3}}{x^{2} + x + 1} d x - \frac{7}{3} \int \frac{d x}{{(x - 1)}^{2}} - \int \frac{d x}{x - 1} =$ $= \frac{17}{6} \int \frac{d x}{x^{2} + x + 1} + \frac{1}{2} \int \frac{2 x + 1}{x^{2} + x + 1} d x - \frac{7}{3} \int \frac{d x}{{(x - 1)}^{2}} - \int \frac{d x}{x - 1} =$ $= \frac{17 \sqrt{3}}{9} \arctan \frac{\sqrt{3} (2 x + 1)}{3} + \frac{1}{2} \ln (x^{2} + x + 1) + \frac{7}{3 (x - 1)} - \ln ∣ x - 1 ∣ =$ $= \frac{17 \sqrt{3}}{9} \arctan \frac{\sqrt{3} (2 x + 1)}{3} + \frac{1}{2} \ln \frac{x^{2} + x + 1}{{(x - 1)}^{2}} + \frac{7}{3 (x - 1)} + C$
	Commented by kaivan.ahmadi last updated on 29/Dec/19

	$t h a n k y o u s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum