∫(√)(x2+2x+2)


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Question Number 54896 by yusufbode1996 last updated on 14/Feb/19

$\int \sqrt{} (x 2 + 2 x + 2)$
Commented by maxmathsup by imad last updated on 14/Feb/19

$l e t I = \int \sqrt{x^{2} + 2 x + 2} d x \Rightarrow I = \int \sqrt{{(x + 1)}^{2} + 1} d x c h a n g e m e n t x + 1 = s h (t) g i v e$ $I = \int c h (t) c h (t) d t = \int \frac{1 + c h (2 t)}{2} d t = \frac{t}{2} + \frac{1}{4} s h (2 t) + c$ $= \frac{t}{2} + \frac{1}{2} s h (t) c h (t) = \frac{1}{2} a r g s h (x + 1) + \frac{1}{2} (x + 1) \sqrt{1 + {(x + 1)}^{2}} + c$ $= \frac{1}{2} l n (x + 1 + \sqrt{x^{2} + 2 x + 2}) + \frac{(x + 1) \sqrt{x^{2} + 2 x + 2}}{2} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19
	$∫(√(x^2 +2x+1+1)) dx ∫(√((x+1)^2 +1))dx now x+1=y 1=a^2 ∫(√(y^2 +a^2 )) dy (√(y^2 +a^2 )) ∫dy−∫[(d/dy)(√(y^2 +a^2 )) ∫dy]dy y(√(y^2 +a^2 )) −∫((2y×y)/(2(√(y^2 +a^2 ))))dy y(√(y^2 +a^2 )) −∫((y^2 +a^2 −a^2 )/(√(y^2 +a^2 )))dy I=y(√(y^2 +a^2 )) −∫(√(y^2 +a^2 )) dy+a^2 ∫(dy/(√(y^2 +a^2 ))) 2I=y(√(y^2 +a^2 )) +a^2 ln(y+(√(y^2 +a^2 )) ) I=(y/2)(√(y^2 +a^2 )) +(a^2 /2)ln(y+(√(y^2 +a^2 )) )+c so required answer is =(((x+1))/2)(√((x+1)^2 +1)) +(1/2)ln{(x+1)+(√((x+1)^2 +1)) }+c$
	$\int \sqrt{x^{2} + 2 x + 1 + 1} d x$ $\int \sqrt{{(x + 1)}^{2} + 1} d x$ $n o w x + 1 = y 1 = a^{2}$ $\int \sqrt{y^{2} + a^{2}} d y$ $\sqrt{y^{2} + a^{2}} \int d y - \int [\frac{d}{d y} \sqrt{y^{2} + a^{2}} \int d y] d y$ $y \sqrt{y^{2} + a^{2}} - \int \frac{2 y \times y}{2 \sqrt{y^{2} + a^{2}}} d y$ $y \sqrt{y^{2} + a^{2}} - \int \frac{y^{2} + a^{2} - a^{2}}{\sqrt{y^{2} + a^{2}}} d y$ $I = y \sqrt{y^{2} + a^{2}} - \int \sqrt{y^{2} + a^{2}} d y + a^{2} \int \frac{d y}{\sqrt{y^{2} + a^{2}}}$ $2 I = y \sqrt{y^{2} + a^{2}} + a^{2} l n (y + \sqrt{y^{2} + a^{2}})$ $I = \frac{y}{2} \sqrt{y^{2} + a^{2}} + \frac{a^{2}}{2} l n (y + \sqrt{y^{2} + a^{2}}) + c$ $s o r e q u i r e d a n s w e r i s$ $= \frac{(x + 1)}{2} \sqrt{{(x + 1)}^{2} + 1} + \frac{1}{2} l n {(x + 1) + \sqrt{{(x + 1)}^{2} + 1}} + c$
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