find-x-1-x-2-1-x-1-x-dx-

Question Number 82440 by mathmax by abdo last updated on 21/Feb/20

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

Commented by abdomathmax last updated on 24/Feb/20

l e t A = \int \frac{x + 1}{x + 2} \sqrt{\frac{1 - x}{1 + x}} d x c h a n g e m e n t \sqrt{\frac{1 - x}{1 + x}} = t

h i v e \frac{1 - x}{1 + x} = t^{2} \Rightarrow 1 - x = t^{2} + {x t}^{2} \Rightarrow 1 - t^{2} = x (1 + t^{2}) \Rightarrow

x = \frac{1 - t^{2}}{1 + t^{2}} \Rightarrow d x = \frac{- 2 t (1 + t^{2}) - (1 - t^{2}) 2 t}{{(1 + t^{2})}^{2}} d t

= \frac{- 2 t - 2 t^{3} - 2 t + 2 t^{3}}{{(1 + t^{2})}^{2}} d t = \frac{- 4 t}{{(1 + t^{2})}^{2}} d t

x + 1 = \frac{1 - t^{2}}{1 + t^{2}} + 1 = \frac{1 - t^{2} + 1 + t^{2}}{1 + t^{2}} = \frac{2}{1 + t^{2}}

x + 2 = \frac{1 - t^{2}}{1 + t^{2}} + 2 = \frac{1 - t^{2} + 2 + 2 t^{2}}{1 + t^{2}} = \frac{3 + t^{2}}{1 + t^{2}}

\Rightarrow I = \int \frac{2}{1 + t^{2}} \times \frac{1 + t^{2}}{3 + t^{2}} \times t \times \frac{- 4 t}{{(1 + t^{2})}^{2}} d t

= \int \frac{- 8 t^{2}}{(t^{2} + 3) {(t^{2} + 1)}^{2}} d t l e t d e c p m p o s e

F (t) = \frac{- 8 t^{2}}{(t^{2} + 3) {(t^{2} + 1)}^{2}} \Rightarrow

F (t) = \frac{a t + b}{t^{2} + 3} + \frac{c t + d}{t^{2} + 1} + \frac{d t + f}{{(t^{2} + 1)}^{2}}

\dots . b e v o n t i n u e d \dots