1-find-dx-x-1-x-2-2-calculate-0-1-dx-x-1-x-2-

Question Number 64418 by turbo msup by abdo last updated on 17/Jul/19

1) f i n d \int \frac{d x}{x + \sqrt{1 + x^{2}}}

2) c a l c u l a t e \int_{0}^{1} \frac{d x}{x + \sqrt{1 + x^{2}}}

Commented by Prithwish sen last updated on 17/Jul/19

a) \int (\sqrt{x^{2} + 1} - x) dx =

\frac{x}{2} \sqrt{x^{2} + 1} + \frac{1}{2} \ln ∣ x + \sqrt{x^{2} + 1} ∣ - \frac{x^{2}}{2} + C

b) {(\frac{x}{2} \sqrt{x^{2} + 1} + \frac{1}{2} \ln ∣ x + \sqrt{x^{2} + 1} ∣ - \frac{x^{2}}{2})}_{0}^{1}

= (\frac{1}{\sqrt{2}} + \frac{1}{2} \ln ∣ 1 + \sqrt{2} ∣ - \frac{1}{2})

Commented by mathmax by abdo last updated on 18/Jul/19

1) l e t I = \int \frac{d x}{x + \sqrt{1 + x^{2}}} \Rightarrow I = \int \frac{x - \sqrt{1 + x^{2}}}{x^{2} - 1 - x^{2}} d x

= \int \sqrt{1 + x^{2}} d x - \int x d x = - \frac{x^{2}}{2} + \int \sqrt{1 + x^{2}} d x + c

c h a n g e m e n t x = s h (t) g i v e

\int \sqrt{1 + x^{2}} d x = \int c h (t) c h t d t = \int {c h}^{2} t d t = \int \frac{1 + c h (2 t)}{2} d t

= \frac{t}{2} + \frac{1}{4} s h (2 t) = \frac{t}{2} + \frac{1}{2} s h (t) c h (t) = \frac{t}{2} + \frac{1}{2} x \sqrt{1 + x^{2}}

= \frac{1}{2} a r g s h (x) + \frac{x \sqrt{1 + x^{2}}}{2} = \frac{1}{2} l n (x + \sqrt{1 + x^{2}}) + \frac{x \sqrt{1 + x^{2}}}{2} \Rightarrow

I = \frac{1}{2} l n (x + \sqrt{1 + x^{2}}) + \frac{x \sqrt{1 + x^{2}}}{2} - \frac{x^{2}}{2} + c .

Commented by mathmax by abdo last updated on 18/Jul/19

2) \int_{0}^{1} \frac{d x}{x + \sqrt{1 + x^{2}}} = \frac{1}{2} {[l n (x + \sqrt{1 + x^{2}}) + x \sqrt{1 + x^{2}} - x^{2}]}_{0}^{1}

= \frac{1}{2} {l n (1 + \sqrt{2}) + \sqrt{2} - 1}