find the value of I= ∫_0 ^1 (dx/((x+1)^2 (√(x^2 +2x +2)))) .


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Question Number 30769 by abdo imad last updated on 25/Feb/18

$f i n d t h e v a l u e o f I = \int_{0}^{1} \frac{d x}{{(x + 1)}^{2} \sqrt{x^{2} + 2 x + 2}} .$
Commented by abdo imad last updated on 27/Feb/18

$w e h a v e x^{2} + 2 x + 2 = {(x + 1)}^{2} + 1 t h e c h = x + 1 = t g i v e$ $I = \int_{1}^{2} \frac{d t}{t^{2} \sqrt{1 + t^{2}}} t h e w e c a n u s e t h e c h . t = t a n θ \Rightarrow$ $I = \int_{\frac{π}{4}}^{a r t a n 2} \frac{1 + {t a n}^{2} θ}{{t a n}^{2} θ \sqrt{1 + {t a n}^{2} θ}} d θ = \int_{\frac{π}{4}}^{a r c t a n 2} \frac{{c o s}^{2} θ}{{s i n}^{2} θ} \frac{1}{c o s θ} d θ$ $= \int_{\frac{π}{4}}^{a r c t a n 2} \frac{c o s θ}{{s i n}^{2} θ} d θ = {[- \frac{1}{s i n θ}]}_{\frac{π}{4}}^{a r c t a n 2}$ $= \sqrt{2} - \frac{1}{s i n (a r c t a n 2)} b u t w e h a v e t h e f o r m u l a$ $s i n (a r c t a n x) = \frac{x}{\sqrt{1 + x^{2}}} (f o r p r o o f p u t x = t a n θ) \Rightarrow$ $s i n (a r c t a n 2) = \frac{2}{\sqrt{5}} \Rightarrow I = \sqrt{2} - \frac{2}{\sqrt{5}} .$
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