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Question Number 12635 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

	Answered by mrW1 last updated on 27/Apr/17

	$x^{2} - 2 x \tan φ - 1$ $= x^{2} - 2 x \tan φ + \tan^{2} φ - 1 - \tan^{2} φ$ $= {(x - \tan φ)}^{2} - {cosec}^{2} φ$ $= (x - \tan φ + cosec φ) (x - \tan φ - cosec φ)$ $\frac{1}{x^{2} - 2 x \tan φ - 1} = \frac{\cos φ}{2} (\frac{1}{x - \tan φ - cosec φ} - \frac{1}{x - \tan φ + cosec φ})$ $\int \frac{d x}{x^{2} - 2 x \tan φ - 1} = \frac{\cos φ}{2} \int (\frac{1}{x - \tan φ - cosec φ} - \frac{1}{x - \tan φ + cosec φ}) d x$ $= \frac{\cos φ}{2} \times \ln ∣ \frac{x - \tan φ - cosec φ}{x - \tan φ + cosec φ} ∣ + C$ $= \frac{\cos φ}{2} \times \ln ∣ \frac{x \cos φ - \sin φ - 1}{x \cos φ - \sin φ + 1} ∣ + C$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

	$t h a k y o u s o m u c h d e a r m r W 1 .$ $d o y o u h a v e a n y i d e a f o r :$ $\int \frac{d x}{x^{2} - 2 x t g φ + 1}$
	Commented by mrW1 last updated on 27/Apr/17
	$x^2 −2xtan ϕ+1 =x^2 −2xtan ϕ+tan^2 ϕ+1−tan^2 ϕ =(x−tan ϕ)^2 +1−tan^2 ϕ ∫(dx/(x^2 −2xtan ϕ+1)) =∫(dx/((x−tan ϕ)^2 +1−tan^2 ϕ)) = { (((1/(√(1−tan^2 ϕ)))tan^(−1) (((x−tan ϕ)/(√(1−tan^2 ϕ))))+C (if 1−tan^2 ϕ>0))),((−(1/(x−tan ϕ)) (if 1−tan^2 ϕ=0))),(((1/(2(√(tan^2 ϕ−1))))ln∣ ((x−tan ϕ−(√(tan^2 ϕ−1)))/(x−tan ϕ+(√(tan^2 ϕ−1))))∣+C (if 1−tan^2 ϕ<0))) :}$
	$x^{2} - 2 x \tan φ + 1$ $= x^{2} - 2 x \tan φ + \tan^{2} φ + 1 - \tan^{2} φ$ $= {(x - \tan φ)}^{2} + 1 - \tan^{2} φ$ $\int \frac{d x}{x^{2} - 2 x \tan φ + 1}$ $= \int \frac{d x}{{(x - \tan φ)}^{2} + 1 - \tan^{2} φ}$ $= {\begin{cases} \frac{1}{\sqrt{1 - \tan^{2} φ}} \tan^{- 1} (\frac{x - \tan φ}{\sqrt{1 - \tan^{2} φ}}) + C (i f 1 - \tan^{2} φ > 0) \\ - \frac{1}{x - \tan φ} (i f 1 - \tan^{2} φ = 0) \\ \frac{1}{2 \sqrt{\tan^{2} φ - 1}} \ln ∣ \frac{x - \tan φ - \sqrt{\tan^{2} φ - 1}}{x - \tan φ + \sqrt{\tan^{2} φ - 1}} ∣ + C (i f 1 - \tan^{2} φ < 0) \end{cases}$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

	$t h a k y o u m y m a s t e r . i t i s p e r f e c t a n d$ $a m a z i n g . g o o d l o c k .$
	Commented by mrW1 last updated on 28/Apr/17

	$t h a n k s, t h e s a m e t o y o u!$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/Apr/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

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