∫ln ∣(√(x+1))+(√x)∣ dx=


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Question Number 40467 by gunawan last updated on 22/Jul/18

$\int \ln ∣ \sqrt{x + 1} + \sqrt{x} ∣ d x =$
Commented by maxmathsup by imad last updated on 22/Jul/18
$let I = ∫ ln∣(√(x+1)) +(√x)∣dx due to x≥0 we have I = ∫ ln((√(x+1))+(√x))dx =_((√x)=t) 2 ∫ln((√(1+t^2 )) +t) t dt by parts u =ln(t+(√(1+t^2 ))) and v^′ =t ⇒ I = 2{ (t^2 /2)ln(t+(√(1+t^2 ))) −∫ (t^2 /2) (1/(√(1+t^2 )))dt} =t^2 ln(t+(√(1+t^2 ))) −∫ ((t^2 +1−1)/(√(1+t^2 )))dt =t^2 ln(t+(√(1+t^2 ))) −∫ (√(1+t^2 ))dt + ∫ (dt/(√(1+t^2 ))) =t^2 ln(t+(√(1+t^2 ))) +ln(t+(√(1+t^2 ))) − ∫(√(1+t^2 ))dt ∫ (√(1+t^2 ))dt =_(t=sh(x)) ∫ch(x)ch(x)= ∫ ((1+ch(2x))/2)dx =(1/2) x +(1/4)sh(2x)=(x/2) +(1/2)sh(x)ch(x) =(1/2)ln(t+(√(1+t^2 ))) +(1/2)t(√(1+t^2 )) ⇒ I = (t^2 +(3/2))ln(t +(√(1+t^2 ))) +((t(√(1+t^2 )))/2) +c .$
$l e t I = \int l n ∣ \sqrt{x + 1} + \sqrt{x} ∣ d x d u e t o x ⩾ 0 w e h a v e$ $I = \int l n (\sqrt{x + 1} + \sqrt{x}) d x =_{\sqrt{x} = t} 2 \int l n (\sqrt{1 + t^{2}} + t) t d t b y p a r t s$ $u = l n (t + \sqrt{1 + t^{2}}) a n d v^{^{'}} = t \Rightarrow$ $I = 2 {\frac{t^{2}}{2} l n (t + \sqrt{1 + t^{2}}) - \int \frac{t^{2}}{2} \frac{1}{\sqrt{1 + t^{2}}} d t}$ $= t^{2} l n (t + \sqrt{1 + t^{2}}) - \int \frac{t^{2} + 1 - 1}{\sqrt{1 + t^{2}}} d t$ $= t^{2} l n (t + \sqrt{1 + t^{2}}) - \int \sqrt{1 + t^{2}} d t + \int \frac{d t}{\sqrt{1 + t^{2}}}$ $= t^{2} l n (t + \sqrt{1 + t^{2}}) + l n (t + \sqrt{1 + t^{2})} - \int \sqrt{1 + t^{2}} d t$ $\int \sqrt{1 + t^{2}} d t =_{t = s h (x)} \int c h (x) c h (x) = \int \frac{1 + c h (2 x)}{2} d x$ $= \frac{1}{2} x + \frac{1}{4} s h (2 x) = \frac{x}{2} + \frac{1}{2} s h (x) c h (x)$ $= \frac{1}{2} l n (t + \sqrt{1 + t^{2})} + \frac{1}{2} t \sqrt{1 + t^{2}} \Rightarrow$ $I = (t^{2} + \frac{3}{2}) l n (t + \sqrt{1 + t^{2}}) + \frac{t \sqrt{1 + t^{2}}}{2} + c .$
Commented by math khazana by abdo last updated on 22/Jul/18

$\Rightarrow I = (x + \frac{3}{2}) l n (\sqrt{x} + \sqrt{1 + x}) + \frac{\sqrt{x} \sqrt{1 + x}}{2} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jul/18
	$use ∫uvdx ln((√(x+1)) +(√x) )x−∫(((1/(2(√(x+1))))+(1/(2(√x) )))/((√(x+1)) +(√x) )).xdx xln((√(x+1)) +(√x) )−(1/2)∫(((√(x+1)) +(√x) )/(((√(x+1)) +(√(x))) (√(x+1)) .(√x)))xdx xln((√(x+1)) +(√x) )−(1/2)∫(((√x) )/(√(x+1)))dx let t^2 =x+1 dx=2tdt I_2 =∫(((√x) )/((√(x+1)) ))dx =∫(((√(t^2 −1)) )/t)2tdt 2∫(√(t^2 −1)) dt 2{(t/2)(√(t^2 −1)) −(1/2)ln(t+(√(t^2 −1)) } {(√(x^2 −1)) .x−ln((√(x^2 −1)) +x)} now rearrannge by yourself...$
	$u s e \int u v d x$ $l n (\sqrt{x + 1} + \sqrt{x}) x - \int \frac{\frac{1}{2 \sqrt{x + 1}} + \frac{1}{2 \sqrt{x}}}{\sqrt{x + 1} + \sqrt{x}} . x d x$ $x l n (\sqrt{x + 1} + \sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x + 1} + \sqrt{x}}{(\sqrt{x + 1} + \sqrt{x)} \sqrt{x + 1} . \sqrt{x}} x d x$ $x l n (\sqrt{x + 1} + \sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{\sqrt{x + 1}} d x$ $l e t t^{2} = x + 1 d x = 2 t d t$ $I_{2} = \int \frac{\sqrt{x}}{\sqrt{x + 1}} d x$ $= \int \frac{\sqrt{t^{2} - 1}}{t} 2 t d t$ $2 \int \sqrt{t^{2} - 1} d t$ $2 {\frac{t}{2} \sqrt{t^{2} - 1} - \frac{1}{2} l n (t + \sqrt{t^{2} - 1}}$ ${\sqrt{x^{2} - 1} . x - l n (\sqrt{x^{2} - 1} + x)}$ $n o w r e a r r a n n g e b y y o u r s e l f . . .$
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