∫((x^2 +3)/(x^6 (x^2 +1)))dx Is there any special method of decomposition other than the use of partial fractions ?


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Question Number 105411 by Ar Brandon last updated on 28/Jul/20
$∫((x^2 +3)/(x^6 (x^2 +1)))dx Is there any special method of decomposition other than the use of partial fractions ?$
$\int \frac{x^{2} + 3}{x^{6} (x^{2} + 1)} d x$ $I s t h e r e a n y s p e c i a l m e t h o d o f d e c o m p o s i t i o n$ $o t h e r t h a n t h e u s e o f p a r t i a l f r a c t i o n s ?$
	Answered by Dwaipayan Shikari last updated on 28/Jul/20

	$\int \frac{x^{2} + 1}{x^{6} (x^{2} + 1)} + \frac{2}{x^{6} (x^{2} + 1)} d x$ $\int \frac{1}{x^{6}} + 2 \int \frac{1}{x^{4}} (\frac{1}{x^{2}} - \frac{1}{x^{2} + 1}) d x$ $3 \int \frac{1}{x^{6}} - 2 \int \frac{1}{x^{4} (x^{2} + 1)} d x$ $3 \int \frac{1}{x^{6}} - 2 \int \frac{1}{x^{2}} (\frac{1}{x^{2}} - \frac{1}{x^{2} + 1}) d x$ $3 \int \frac{1}{x^{6}} - 2 \int \frac{1}{x^{4}} + 2 \int \frac{1}{x^{2} (x^{2} + 1)} d x$ $3 \int \frac{1}{x^{6}} - 2 \int \frac{1}{x^{4}} + 2 \int \frac{1}{x^{2}} - 2 \int \frac{1}{x^{2} + 1} d x$ $- \frac{3}{5 x^{5}} + \frac{2}{3 x^{3}} - \frac{2}{x} - 2 {t a n}^{- 1} x + C$
	Commented by Dwaipayan Shikari last updated on 28/Jul/20

	$I t h i n k s o . I s i t a b e t t e r w a y t h a t y o u h a v e m e n t i o n e d ?$
	Commented by Dwaipayan Shikari last updated on 28/Jul/20
	��
	Commented by Ar Brandon last updated on 28/Jul/20
	Perfect, bro. Exactly what I needed. Thanks��
	Commented by Ar Brandon last updated on 28/Jul/20

	$\frac{1}{x^{2} (x^{2} + 1)} = \frac{x^{2} + 1 - x^{2}}{x^{2} (x^{2} + 1)} = \frac{x^{2} + 1}{x^{2} (x^{2} + 1)} - \frac{x^{2}}{x^{2} (x^{2} + 1)}$ $= \frac{1}{x^{2}} - \frac{1}{x^{2} + 1},$
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