Evaluate ∫((x^4 +1)/(x^6 +1))dx [W.B.H.S 2018]


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 32951 by math1967 last updated on 07/Apr/18

$E v a l u a t e$ $\int \frac{x^{4} + 1}{x^{6} + 1} d x [W . B . H . S 2018]$
Commented by abdo imad last updated on 09/Apr/18

$y e s i h a v e c o m m i t e d a e r r o r b e c a u s i d o n t g i v e$ $a t t e n t i o n t o l i m i t s (b o r n e s) t h a n k f o r t h i s r e m a r k . . .$ $i h a v e d e l e t e d t h e p o s t . .$
	Answered by Joel578 last updated on 07/Apr/18

	$I = \int \frac{x^{4} + 1}{x^{6} + 1} d x = \int \frac{x^{4} + 1 + x^{2} - x^{2}}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x$ $= \int \frac{x^{4} - x^{2} + 1}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x + \int \frac{x^{2}}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x$ $= \int \frac{1}{x^{2} + 1} d x + \int \frac{x^{2}}{x^{6} + 1} d x$ $I_{2} = \int \frac{x^{2}}{x^{6} + 1} d x (u = x^{3} \to d u = 3 x^{2} d x)$ $= \int \frac{x^{2}}{u^{2} + 1} (\frac{d u}{3 x^{2}}) = \frac{1}{3} \int \frac{d u}{u^{2} + 1}$ $= \frac{1}{3} \tan^{- 1} (u) + C$ $I = \int \frac{1}{x^{2} + 1} d x + \int \frac{x^{2}}{x^{6} + 1} d x$ $= \tan^{- 1} (x) + \frac{1}{3} \tan^{- 1} (u) + C$ $= \tan^{- 1} (x) + \frac{1}{3} \tan^{- 1} (x^{3}) + C$
	Answered by MJS last updated on 07/Apr/18

	$lol! I did it the hard way . . .$ $x^{6} + 1 = (x^{4} - x^{2} + 1) (x^{2} + 1) =$ $= (x^{2} - \sqrt{3} x + 1) (x^{2} + \sqrt{3} x + 1) (x^{2} + 1)$ $\frac{x^{4} + 1}{x^{6} + 1} = \frac{N_{1}}{x^{2} - \sqrt{3} x + 1} + \frac{N_{2}}{x^{2} + \sqrt{3} + 1} + \frac{N_{3}}{x^{2} + 1}$ $N_{1} = N_{2} = \frac{1}{6}; N_{3} = \frac{2}{3}$ $\int \frac{x^{4} + 1}{x^{6} + 1} d x = \int \frac{1}{6 (x^{2} - \sqrt{3} x + 1)} d x + \int \frac{1}{6 (x^{2} + \sqrt{3} x + 1)} d x + \int \frac{2}{3 (x^{2} + 1)} d x =$ $[\int \frac{1}{{a x}^{2} + b x + c} d x = \frac{2}{\sqrt{- b^{2} + 4 a c}} \tan^{- 1} \frac{2 a x + b}{\sqrt{- b^{2} + 4 a c}}; b^{2} - 4 a c < 0]$ $= \frac{1}{3} (\tan^{- 1} (2 x - \sqrt{3}) + \tan^{- 1} (2 x + \sqrt{3}) + {2 \tan}^{- 1} (x)) + C$
	Commented by Joel578 last updated on 08/Apr/18

	$Great work Sir$
	Commented by MJS last updated on 08/Apr/18

	$thanks, but I like your solution$ $more than mine$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum