find-the-value-of-I-0-1-dx-x-1-2-x-2-2x-2-

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}} .