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Question Number 85355 by TawaTawa1 last updated on 21/Mar/20

Commented by mathmax by abdo last updated on 21/Mar/20
$I =∫ arctan((√(1+(√x))))dx changement (√(1+(√x)))=t give 1+(√x)=t^2 ⇒(√x)=t^2 −1 ⇒x =(t^2 −1)^2 ⇒dx =2(2t)(t^2 −1) ⇒ I =∫ arctan(t)(4t(t^2 −1))dt =4 ∫ (t^3 −t) arctant dt =_(byparts) 4{ (t^4 /4)−(t^2 /2) −∫ ((t^4 /4)−(t^2 /2))(dt/(1+t^2 ))} =t^4 −2t^2 −∫ ((t^4 −t^2 )/(t^2 +1))dt we have ∫ ((t^4 −t^2 )/(t^2 +1))dt =∫ ((t^2 (t^2 +1−2))/(t^2 +1))dt =∫ t^2 dt −∫ ((2t^2 )/(t^2 +1))dt =(t^3 /3)−2 ∫ ((t^2 +1−1)/(t^2 +1))dt =(t^3 /3)−2t +2arctan(t) +c ⇒ I =((√(1+(√x))))^4 −2((√(1+(√x))))^2 −(1/3)((√(1+(√x))))^3 +2(√(1+(√x))) +2arctan((√(1+(√x)))) +c$
$I = \int a r c t a n (\sqrt{1 + \sqrt{x}}) d x c h a n g e m e n t \sqrt{1 + \sqrt{x}} = t g i v e$ $1 + \sqrt{x} = t^{2} \Rightarrow \sqrt{x} = t^{2} - 1 \Rightarrow x = {(t^{2} - 1)}^{2} \Rightarrow d x = 2 (2 t) (t^{2} - 1) \Rightarrow$ $I = \int a r c t a n (t) (4 t (t^{2} - 1)) d t$ $= 4 \int (t^{3} - t) a r c t a n t d t =_{b y p a r t s} 4 {\frac{t^{4}}{4} - \frac{t^{2}}{2} - \int (\frac{t^{4}}{4} - \frac{t^{2}}{2}) \frac{d t}{1 + t^{2}}}$ $= t^{4} - 2 t^{2} - \int \frac{t^{4} - t^{2}}{t^{2} + 1} d t w e h a v e$ $\int \frac{t^{4} - t^{2}}{t^{2} + 1} d t = \int \frac{t^{2} (t^{2} + 1 - 2)}{t^{2} + 1} d t = \int t^{2} d t - \int \frac{2 t^{2}}{t^{2} + 1} d t$ $= \frac{t^{3}}{3} - 2 \int \frac{t^{2} + 1 - 1}{t^{2} + 1} d t = \frac{t^{3}}{3} - 2 t + 2 a r c t a n (t) + c \Rightarrow$ $I = {(\sqrt{1 + \sqrt{x}})}^{4} - 2 {(\sqrt{1 + \sqrt{x}})}^{2} - \frac{1}{3} {(\sqrt{1 + \sqrt{x}})}^{3} + 2 \sqrt{1 + \sqrt{x}} + 2 a r c t a n (\sqrt{1 + \sqrt{x}}) + c$
Commented by mathmax by abdo last updated on 21/Mar/20

$e r r o r a t l i n e 5 I = t^{4} - 2 t^{2} - \int \frac{t^{4} - 2 t^{2}}{t^{2} + 1} d t = . . . .$
Commented by TawaTawa1 last updated on 22/Mar/20

$God bless you sir .$
Commented by abdomathmax last updated on 23/Mar/20

$y o u a r e w e l c o m e m i s s .$
	Answered by mind is power last updated on 21/Mar/20

	$b y p a r t$ $\int {t a n}^{-} (\sqrt{1 + \sqrt{x}}) d x = (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - \int (x - 4) . \frac{1}{4 \sqrt{x} . \sqrt{1 + \sqrt{x}}} . \frac{d x}{(2 + \sqrt{x})}$ $= (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - \int \frac{\sqrt{x} - 2}{4 \sqrt{x} . \sqrt{1 + \sqrt{x}}} d x$ $u = \sqrt{x} \Rightarrow d u = \frac{d x}{2 \sqrt{x}}$ $\Rightarrow (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - \int \frac{u - 2}{2 \sqrt{1 + u}} d u$ $= (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - \int \frac{\sqrt{u + 1}}{2} d u + \int \frac{3 d u}{2 \sqrt{1 + u}}$ $= (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - \frac{1}{3} {(u + 1)}^{\frac{3}{2}} + 3 \sqrt{1 + u} + c$ $= (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) + (\frac{- (u + 1)}{3} + 3) \sqrt{1 + u} + c$ $u = \sqrt{x}$ $(x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) + (- \frac{\sqrt{x}}{3} + \frac{8}{3}) \sqrt{1 + \sqrt{x}} + c$ $= (x - 4) {t a n}^{-} (\sqrt{1 + \sqrt{x}}) - (\frac{\sqrt{x} - 8}{3}) \sqrt{1 + \sqrt{x}} + c$
	Commented by TawaTawa1 last updated on 21/Mar/20

	$God bless you sir, i appreciate your time .$
	Commented by mind is power last updated on 23/Mar/20

	$w i t h e P l e a s u r m i s s$
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