Integrate..∫(√(1+(√(1+(√x))))) dx


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Question Number 55873 by Easyman32 last updated on 05/Mar/19

$I n t e g r a t e . . \int \sqrt{1 + \sqrt{1 + \sqrt{x}}} d x$
Commented by maxmathsup by imad last updated on 05/Mar/19
$let I =∫(√(1+(√(1+(√x))))) dx changement (√(1+(√x)))=t give 1+(√x)=t^2 ⇒ (√x)=t^2 −1 ⇒x =(t^2 −1)^2 =t^4 −2t^2 +1 ⇒ I =∫ (√(1+t))(4t^3 −4t)dt =4 ∫ (t^3 −t)(√(1+t))dt =_((√(1+t))=u) 4∫ {(u^2 −1)^3 −(u^2 −1))u(2u)du =8 ∫ (u^2 −1){ (u^2 −1)^2 −1}u^2 du =8 ∫ (u^4 −u^2 )( u^4 −2u^2 )du =8 ∫ (u^8 −3u^6 +2u^4 }du =8{(1/9)u^9 −(3/7)u^7 +(2/5)u^5 } +c =(8/9)u^9 −((24)/7)u^7 +((16)/5)u^5 +c =(8/9)((√(1+t)))^9 −((24)/7)((√(1+t)))^7 +((16)/5)((√(1+t)))^5 +c =(8/9)((√(1+(√(1+(√x))))))^9 −((24)/7)((√(1+(√(1+(√x))))))^7 +((16)/5)((√(1+(√(1+(√x))))))^5 +c .$
$l e t I = \int \sqrt{1 + \sqrt{1 + \sqrt{x}}} d x c h a n g e m e n t \sqrt{1 + \sqrt{x}} = t g i v e 1 + \sqrt{x} = t^{2} \Rightarrow$ $\sqrt{x} = t^{2} - 1 \Rightarrow x = {(t^{2} - 1)}^{2} = t^{4} - 2 t^{2} + 1 \Rightarrow$ $I = \int \sqrt{1 + t} (4 t^{3} - 4 t) d t = 4 \int (t^{3} - t) \sqrt{1 + t} d t =_{\sqrt{1 + t} = u} 4 \int {{(u^{2} - 1)}^{3} - (u^{2} - 1)) u (2 u) d u$ $= 8 \int (u^{2} - 1) {{(u^{2} - 1)}^{2} - 1} u^{2} d u$ $= 8 \int (u^{4} - u^{2}) (u^{4} - 2 u^{2}) d u = 8 \int (u^{8} - 3 u^{6} + 2 u^{4}} d u$ $= 8 {\frac{1}{9} u^{9} - \frac{3}{7} u^{7} + \frac{2}{5} u^{5}} + c$ $= \frac{8}{9} u^{9} - \frac{24}{7} u^{7} + \frac{16}{5} u^{5} + c = \frac{8}{9} {(\sqrt{1 + t})}^{9} - \frac{24}{7} {(\sqrt{1 + t})}^{7} + \frac{16}{5} {(\sqrt{1 + t})}^{5} + c$ $= \frac{8}{9} {(\sqrt{1 + \sqrt{1 + \sqrt{x}}})}^{9} - \frac{24}{7} {(\sqrt{1 + \sqrt{1 + \sqrt{x}}})}^{7} + \frac{16}{5} {(\sqrt{1 + \sqrt{1 + \sqrt{x}}})}^{5} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19
	$x=t^2 dx=2tdt ∫(√(1+(√(1+t)))) ×2tdt k^2 =1+t 2kdk=dt ∫(√(1+k)) ×2(k^2 −1)×2kdk 4∫(√(1+k)) ×(k^3 −k)dk a^2 =1+k 2ada=dk 4∫a×{(a^2 −1)^3 −(a^2 −1)}×2ada 8∫a^2 (a^6 −3a^4 +3a^2 −1−a^2 +1) 8∫a^8 −3a^6 +2a^4 da 8×{(a^9 /9)−((3a^7 )/7)+((2a^5 )/5)}+c wait pls busy ... 8×{(((1+k)^(9/2) )/9)−(3/7)(1+k)^(7/2) +(2/5)(1+k)^(5/2) }+c k=1+t t=(√x) so k=1+(√x) 8×{(((2+(√x) )^(9/2) )/9)−(3/7)(2+(√x) )^(7/2) +(2/5)(2+(√x) )^(5/2) }+c pls check...$
	$x = t^{2} d x = 2 t d t$ $\int \sqrt{1 + \sqrt{1 + t}} \times 2 t d t$ $k^{2} = 1 + t 2 k d k = d t$ $\int \sqrt{1 + k} \times 2 (k^{2} - 1) \times 2 k d k$ $4 \int \sqrt{1 + k} \times (k^{3} - k) d k$ $a^{2} = 1 + k 2 a d a = d k$ $4 \int a \times {{(a^{2} - 1)}^{3} - (a^{2} - 1)} \times 2 a d a$ $8 \int a^{2} (a^{6} - 3 a^{4} + 3 a^{2} - 1 - a^{2} + 1)$ $8 \int a^{8} - 3 a^{6} + 2 a^{4} d a$ $8 \times {\frac{a^{9}}{9} - \frac{3 a^{7}}{7} + \frac{2 a^{5}}{5}} + c$ $w a i t p l s b u s y . . .$ $8 \times {\frac{{(1 + k)}^{\frac{9}{2}}}{9} - \frac{3}{7} {(1 + k)}^{\frac{7}{2}} + \frac{2}{5} {(1 + k)}^{\frac{5}{2}}} + c$ $k = 1 + t t = \sqrt{x} s o k = 1 + \sqrt{x}$ $8 \times {\frac{{(2 + \sqrt{x})}^{\frac{9}{2}}}{9} - \frac{3}{7} {(2 + \sqrt{x})}^{\frac{7}{2}} + \frac{2}{5} {(2 + \sqrt{x})}^{\frac{5}{2}}} + c$ $p l s c h e c k . . .$
	Answered by MJS last updated on 05/Mar/19

	$\int \sqrt{1 + \sqrt{1 + \sqrt{x}}} d x =$ $[t = 1 + \sqrt{1 + \sqrt{x}} \to d x = 4 \sqrt{x} \sqrt{1 + \sqrt{x}} d t; \sqrt{x} = t^{2} - 2 t]$ $= 4 \int (t - 2) (t - 1) t^{\frac{3}{2}} d t =$ $= 4 \int (t^{\frac{7}{2}} - 3 t^{\frac{5}{2}} + 2 t^{\frac{3}{2}}) d t =$ $= \frac{8}{9} t^{\frac{9}{2}} - \frac{24}{7} t^{\frac{7}{2}} + \frac{16}{5} t^{\frac{5}{2}} =$ $= \frac{8}{315} t^{\frac{5}{2}} (35 t^{2} - 135 t + 126)$ $please do the inserting . . .$
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