∫ (dx/( (√x)+x(√(x+1)))) =?


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Question Number 157309 by cortano last updated on 22/Oct/21

$\int \frac{d x}{\sqrt{x} + x \sqrt{x + 1}} = ?$
	Answered by MJS_new last updated on 22/Oct/21

	$\int \frac{d x}{\sqrt{x} + x \sqrt{x + 1}} =$ $= \int \frac{\sqrt{x + 1}}{x^{2} + x - 1} d x - \int \frac{d x}{(x^{2} + x - 1) \sqrt{x}}$ $\int \frac{\sqrt{x + 1}}{x^{2} + x - 1} d x =$ $[t = \sqrt{x + 1} \to d x = 2 \sqrt{x + 1} d t]$ $= 2 \int \frac{t^{2}}{t^{4} - t^{2} + 1} d t$ $. . .$ $- \int \frac{d x}{(x^{2} + x - 1) \sqrt{x}} =$ $[u = \sqrt{x} \to d x = 2 \sqrt{x} d u]$ $= - 2 \int \frac{d u}{u^{4} + u^{2} - 1}$ $. . .$
	Commented by lyubita last updated on 22/Oct/21

	$n e x t t i m e d o n o t a n s w e r {c o r t a n o}^{'} s$ $q u e s t i o n$
	Commented by cortano last updated on 22/Oct/21

	$a r e y o u c r a z y l u b i t o n g$
	Answered by MJS_new last updated on 22/Oct/21

	$\int \frac{d x}{\sqrt{x} + x \sqrt{x + 1}} =$ $[t = \sqrt{x} + \sqrt{x + 1} \to d x = \frac{2 \sqrt{x} \sqrt{x + 1}}{\sqrt{x} + \sqrt{x + 1}} d t]$ $= 4 \int \frac{t^{2} + 1}{t^{4} + 4 t^{2} - 1} d t$ $. . .$
	Commented by cortano last updated on 22/Oct/21

	$h a h a s u p e r b e a s y$
	Commented by yeti123 last updated on 22/Oct/21

	$cortano, you said super easy, but you {can}^{'} t solve .$
	Commented by cortano last updated on 22/Oct/21
	hey how do you know i can't finish it? are you God?
	Commented by MJS_new last updated on 22/Oct/21

	$he {can}^{'} t be god because I am god and {there}^{'} s$ $only one god$
	Commented by ajfour last updated on 22/Oct/21

	$w h y d i d u h a v e t o d i v u l g e t h a t$ $s o s o o n, w e h a d p l a n n e d t o$ $k e e p h i m f o o l i n r o u n d a l i t t l e$ $l o n g e r . .$
	Commented by MJS_new last updated on 22/Oct/21

	$just saw your other comment, good song .$ $‘ ‘ {alphaprime}^{″} and his mates vanished it$ $seems . I found no trace of them on the www$
	Commented by ajfour last updated on 22/Oct/21
	please check alphaprime.slack.com
	Commented by cortano last updated on 22/Oct/21

	$h a h a . . i t t r u e p r o f f m j s s i r . i b e l i e v e d$ $i t . y o u a r e a e x c e l e n t .$
	Commented by talminator2856791 last updated on 22/Oct/21

	$it sound like you all know each other$ $personally .$
	Commented by cortano last updated on 23/Oct/21
	$I=4∫ ((t^2 +1)/((t^2 +2)^2 −5)) dt I=4∫ ((t^2 +1)/((t^2 +2+(√5))(t^2 +2−(√5)))) dt I=4∫ {(((1/(10))((√5)+5))/(t^2 +((√5)+2)))+(((1/(10))((√5)−5))/(t^2 −((√5)−2)))}dt I=((2((√5)+5))/(5(√((√5)+2)))) arctan ((t/( (√((√5)−2)))))−((2((√5)−5))/(5(√((√5)−2))))tanh^(−1) ((t/( (√((√5)−2)))))+C where t=(√(x+1))+(√x)$
	$I = 4 \int \frac{t^{2} + 1}{{(t^{2} + 2)}^{2} - 5} d t$ $I = 4 \int \frac{t^{2} + 1}{(t^{2} + 2 + \sqrt{5}) (t^{2} + 2 - \sqrt{5})} d t$ $I = 4 \int {\frac{\frac{1}{10} (\sqrt{5} + 5)}{t^{2} + (\sqrt{5} + 2)} + \frac{\frac{1}{10} (\sqrt{5} - 5)}{t^{2} - (\sqrt{5} - 2)}} d t$ $I = \frac{2 (\sqrt{5} + 5)}{5 \sqrt{\sqrt{5} + 2}} \arctan (\frac{t}{\sqrt{\sqrt{5} - 2}}) - \frac{2 (\sqrt{5} - 5)}{5 \sqrt{\sqrt{5} - 2}} \tanh^{- 1} (\frac{t}{\sqrt{\sqrt{5} - 2}}) + C$ $w h e r e t = \sqrt{x + 1} + \sqrt{x}$
	Answered by ajfour last updated on 22/Oct/21
	$I=2∫((dx/(2(√x)))/(1+(√x)(√(x+1)))) let (√x)=t ⇒ (dx/(2(√x)))=dt I=2∫(dt/(1+t(√(t^2 +1)))) let t=tan θ I=2∫((sec^2 θdθ)/(1+sec θtan θ)) I=2∫(dθ/(cos^2 θ+sin θ)) I=−2∫(dθ/(sin^2 θ−sin θ+1)) (1/(s^2 −s+1))=(1/(ω(ω−1))){(1/(s+ω))−(1/(s+ω^2 ))} I=(2/(ω(1−ω))){∫(dθ/(sin θ+ω))−∫(dθ/(sin θ+ω^2 ))} now its a piece of cake, just for cortano sir.$
	$I = 2 \int \frac{\frac{d x}{2 \sqrt{x}}}{1 + \sqrt{x} \sqrt{x + 1}}$ $l e t \sqrt{x} = t \Rightarrow \frac{d x}{2 \sqrt{x}} = d t$ $I = 2 \int \frac{d t}{1 + t \sqrt{t^{2} + 1}}$ $l e t t = \tan θ$ $I = 2 \int \frac{\sec^{2} θ d θ}{1 + \sec θ \tan θ}$ $I = 2 \int \frac{d θ}{\cos^{2} θ + \sin θ}$ $I = - 2 \int \frac{d θ}{\sin^{2} θ - \sin θ + 1}$ $\frac{1}{s^{2} - s + 1} = \frac{1}{ω (ω - 1)} {\frac{1}{s + ω} - \frac{1}{s + ω^{2}}}$ $I = \frac{2}{ω (1 - ω)} {\int \frac{d θ}{\sin θ + ω} - \int \frac{d θ}{\sin θ + ω^{2}}}$ $n o w i t s a p i e c e o f c a k e,$ $j u s t f o r c o r t a n o s i r .$
	Commented by cortano last updated on 22/Oct/21

	$w a w {t h a t}^{'} s s o s w e e t s i r a j f o u r$
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