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Question Number 41436 by Raj Singh last updated on 07/Aug/18

	Answered by rahul 19 last updated on 07/Aug/18

	$Check Q . Id 36491$
	Answered by MJS last updated on 07/Aug/18

	$\int \frac{d x}{x^{4} + 8 x^{2} + 9} = \int \frac{d x}{(x^{2} + 4 - \sqrt{7}) (x^{2} + 4 + \sqrt{7})} =$ $\frac{\sqrt{7}}{14} \int \frac{d x}{x^{2} + 4 - \sqrt{7}} - \frac{\sqrt{7}}{14} \int \frac{d x}{x^{2} + 4 + \sqrt{7}} =$ $[\int \frac{d t}{t^{2} + a} = \frac{\sqrt{a}}{a} \arctan \frac{x \sqrt{a}}{a}]$ $= (\frac{7 \sqrt{2} + \sqrt{14}}{84}) \arctan (\frac{\sqrt{14} + \sqrt{2}}{6} x) - (\frac{7 \sqrt{2} - \sqrt{14}}{84}) \arctan (\frac{\sqrt{14} - \sqrt{2}}{6} x) + 4 c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
	$1)∫(dx/(x^4 +8x^2 +9)) =∫(dx/((x^2 )^2 +2.x^2 .4+16−7)) =∫(dx/((x^2 +4)^2 −((√7))^2 )) =∫(dx/((x^2 +4+(√(7 )) )(x^2 +4−(√7) ))) = (1/(2(√7))){∫(dx/(x^2 +4−(√7)))−∫(dx/(x^2 +4+(√7)))} =(1/(2(√7))){(1/(√(4−(√7))))tan^(−1) ((x/(√(4−(√7)))))−(1/(√(4+(√7))))tan^(−1) ((x/(√(4+(√7)))))$
	$1) \int \frac{d x}{x^{4} + 8 x^{2} + 9}$ $= \int \frac{d x}{{(x^{2})}^{2} + 2 . x^{2} . 4 + 16 - 7}$ $= \int \frac{d x}{{(x^{2} + 4)}^{2} - {(\sqrt{7})}^{2}}$ $= \int \frac{d x}{(x^{2} + 4 + \sqrt{7}) (x^{2} + 4 - \sqrt{7})}$ $= \frac{1}{2 \sqrt{7}} {\int \frac{d x}{x^{2} + 4 - \sqrt{7}} - \int \frac{d x}{x^{2} + 4 + \sqrt{7}}}$ $= \frac{1}{2 \sqrt{7}} {\frac{1}{\sqrt{4 - \sqrt{7}}} {t a n}^{- 1} (\frac{x}{\sqrt{4 - \sqrt{7}}}) - \frac{1}{\sqrt{4 + \sqrt{7}}} {t a n}^{- 1} (\frac{x}{\sqrt{4 + \sqrt{7}}})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

	$2) t^{2} = s e c x + t a n x$ $2 t d t = s e c x t a n x + {s e c}^{2} x d x = s e c x (s e c x + t a n x) d x$ $2 t d t = s e c x (t^{2}) d x$ $\frac{2 d t}{t} = s e c x d x$ ${s e c}^{2} x - {t a n}^{2} x = 1$ $(s e c x + t a n x) (s e c x - t a n x) = 1$ $s e c x - t a n x = \frac{1}{t^{2}}$ $s e c x + t a n x = t^{2}$ $s e c x = \frac{1}{2} (t^{2} + \frac{1}{t^{2}})$ $\int \frac{{s e c}^{2} x}{{(s e c x + t a n x)}^{\frac{9}{2}}} d x$ $\int \frac{s e c x . s e c x d x}{{(s e c x + t a n x)}^{\frac{9}{2}}}$ $\int \frac{\frac{1}{2} (t^{2} + \frac{1}{t^{2}}) \times \frac{2 d t}{t}}{{(t^{2})}^{\frac{9}{2}}}$ $\int \frac{t^{2} + \frac{1}{t^{2}}}{t^{10}} d t$ $\int (t^{- 8} + t^{- 12}) d t$ $= \frac{t^{- 7}}{- 7} + \frac{t^{- 11}}{- 11} c$ $= \frac{{(s e c x + t a n x)}^{\frac{- 7}{2}}}{- 7} + \frac{{(s e c x + t a n x)}^{\frac{- 11}{2}}}{- 11} + c$
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