1-2-1-2-x-2-1-x-4-x-2-1-dx-

Question Number 192712 by cortano12 last updated on 25/May/23

\underset{- 1 / 2}{\int^{1 / 2}} \sqrt{x^{2} + 1 + \sqrt{x^{4} + x^{2} + 1}} dx = ?

Answered by horsebrand11 last updated on 25/May/23

I = 2 \underset{0}{\int^{1 / 2}} \sqrt{x^{2} + 1 + \sqrt{(x^{2} + x + 1) (x^{2} - x + 1)}} d x

= \sqrt{2} \underset{0}{\int^{1 / 2}} \sqrt{{(\sqrt{x^{2} + x + 1}) + \sqrt{x^{2} - x + 1})}^{2}} d x

= \sqrt{2} \underset{0}{\int^{1 / 2}} (\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x + 1}) d x

= \frac{1}{\sqrt{2}} (\underset{0}{\int^{1 / 2}} \sqrt{{(2 x + 1)}^{2} + 3} d x + \underset{- 1 / 2}{\int^{0}} \sqrt{{(2 x + 1)}^{2} + 3} d x)

= \frac{1}{\sqrt{2}} \underset{- 1 / 2}{\int^{1 / 2}} \sqrt{{(2 x + 1)}^{2} + 3} d x

l e t 2 x + 1 = \sqrt{3} \tan α

I = \frac{3}{2 \sqrt{2}} \underset{0}{\int^{\arctan (2 / \sqrt{3)}}} \sec^{3} α d α

= \frac{3}{4 \sqrt{2}} {(\sec α \tan α + \ln (\sec α + \tan α)]}_{0}^{\arctan (\frac{2}{\sqrt{3}})}

= 2 \sqrt{7} + 3 \ln (\frac{2 + \sqrt{7}}{\sqrt{3}})