∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx


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Question Number 87686 by M±th+et£s last updated on 05/Apr/20

$\int \sqrt{\frac{l n (x + \sqrt{1 + x^{2}})}{1 + x^{2}}} d x$
	Answered by TANMAY PANACEA. last updated on 05/Apr/20
	$t^2 =ln(x+(√(1+x^2 )) ) 2t(dt/dx)=(1/(x+(√(1+x^2 ))))×(1+((2x)/(2(√(1+x^2 )))))=(1/(√(1+x^2 ))) 2tdt=(dx/(√(1+x^2 ))) ∫(√(ln(x+(√(1+x^2 )) )) ×(dx/(√(1+x^2 ))) ∫t×2tdt 2×(t^3 /3)+c (2/3){ln(x+(√(1+x^2 )) )}^(3/2) +c$
	$t^{2} = l n (x + \sqrt{1 + x^{2}})$ $2 t \frac{d t}{d x} = \frac{1}{x + \sqrt{1 + x^{2}}} \times (1 + \frac{2 x}{2 \sqrt{1 + x^{2}}}) = \frac{1}{\sqrt{1 + x^{2}}}$ $2 t d t = \frac{d x}{\sqrt{1 + x^{2}}}$ $\int \sqrt{l n (x + \sqrt{1 + x^{2}}} \times \frac{d x}{\sqrt{1 + x^{2}}}$ $\int t \times 2 t d t$ $2 \times \frac{t^{3}}{3} + c$ $\frac{2}{3} {l n (x + \sqrt{1 + x^{2}})}^{\frac{3}{2}} + c$
	Answered by Ar Brandon last updated on 05/Apr/20
	$I=∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx Let x=tanθ⇒dx=sec^2 θdθ ⇒I=∫(√(ln(sec θ+tan θ))) sec θdθ ⇒I=(√(ln(sec θ+tan θ)))∫sec θ−∫{(d/dθ)(√(ln(sec θ+tan θ)))∫sec θ}dθ ⇒I=(ln(sec θ+tan θ))^(3/2) −∫{((sec θ)/(2(√(ln(sec θ+tan θ)))))∙ln(sec θ+tan θ)}dθ ⇒I=(ln(sec θ+tan θ))^(3/2) −(1/2)∫(√(ln(sec θ+tan θ))) secdθ ⇒(3/2)I=(ln(sec θ+tan θ))^(3/2) ⇒I=(2/3)(ln(sec θ+tan θ))^(3/2) ⇒∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx=(2/3)(ln(sec(tan^(−1) x)+x))^(3/2) +C C∈R$
	$I = \int \sqrt{\frac{l n (x + \sqrt{1 + x^{2}})}{1 + x^{2}}} d x$ $L e t x = t a n θ \Rightarrow d x = {s e c}^{2} θ d θ$ $\Rightarrow I = \int \sqrt{l n (s e c θ + t a n θ)} s e c θ d θ$ $\Rightarrow I = \sqrt{l n (s e c θ + t a n θ)} \int s e c θ - \int {\frac{d}{d θ} \sqrt{l n (s e c θ + t a n θ)} \int s e c θ} d θ$ $\Rightarrow I = {(l n (s e c θ + t a n θ))}^{\frac{3}{2}} - \int {\frac{s e c θ}{2 \sqrt{l n (s e c θ + t a n θ)}} \cdot l n (s e c θ + t a n θ)} d θ$ $\Rightarrow I = {(l n (s e c θ + t a n θ))}^{\frac{3}{2}} - \frac{1}{2} \int \sqrt{l n (s e c θ + t a n θ)} s e c d θ$ $\Rightarrow \frac{3}{2} I = {(l n (s e c θ + t a n θ))}^{\frac{3}{2}}$ $\Rightarrow I = \frac{2}{3} {(l n (s e c θ + t a n θ))}^{\frac{3}{2}}$ $\Rightarrow \int \sqrt{\frac{l n (x + \sqrt{1 + x^{2}})}{1 + x^{2}}} d x = \frac{2}{3} {(l n (s e c ({t a n}^{- 1} x) + x))}^{\frac{3}{2}} + C$ $C \in R$
	Commented by M±th+et£s last updated on 05/Apr/20

	$g o d b l e s s y o u$
	Commented by Ar Brandon last updated on 05/Apr/20
	*G. ��
	Answered by mind is power last updated on 05/Apr/20

	$x = s h (t) \Rightarrow$ $= \int \sqrt{\frac{l n (s h (t) + c h (t))}{1 + {s h}^{2} (t)}} . c h (t) d t$ $= \int \sqrt{\frac{l n (e^{t})}{{c h}^{2} (t)}} . c h (t) d t = \int \sqrt{t} d t = \frac{3}{2} t^{\frac{3}{2}} + c$ $= \frac{3}{2} {a r g s h}^{\frac{3}{2}} (x) + c$
	Commented by M±th+et£s last updated on 05/Apr/20

	$g o d b l e s s y o u s i r$
	Commented by Ar Brandon last updated on 05/Apr/20

	$I t h i n k i t s h o u l d b e \frac{2}{3}$
	Commented by M±th+et£s last updated on 05/Apr/20

	$y e s i t h i n k i t s t y p o$
	Commented by Ar Brandon last updated on 05/Apr/20

	$N i c e i n i t i a t i v e t h o u g h . B r a v o!$
	Commented by mind is power last updated on 06/Apr/20

	$y e a h s o r r h \frac{2}{3}$
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