calculate-1-x-2-2x-1-x-2-2x-dx-

Question Number 60687 by maxmathsup by imad last updated on 24/May/19

c a l c u l a t e \int \frac{\sqrt{1 + x^{2}} - 2 x}{\sqrt{1 + x^{2}} + 2 x} d x

Commented by maxmathsup by imad last updated on 25/May/19

l e t A = \int \frac{\sqrt{1 + x^{2}} - 2 x}{\sqrt{1 + x^{2}} + 2 x} d x l e t u s e t h e c h a n g e m e n t x = s h (t) \Rightarrow

A = \int \frac{c h (t) - 2 s h (t)}{c h (t) + 2 s h (t)} c h (t) d t = \int \frac{{c h}^{2} t - s h (2 t)}{c h (t) + 2 s h (t)} d t

= \int \frac{\frac{1 + c h (2 t)}{2} - s h (2 t)}{c h (t) + 2 s h (t)} d t = \frac{1}{2} \int \frac{1 + c h (2 t) - 2 s h (2 t)}{c h (t) + 2 s h (t)} d t

2 A = \int \frac{1 + \frac{e^{2 t} + e^{- 2 t}}{2} - 2 \frac{e^{2 t} - e^{- 2 t}}{2}}{\frac{e^{t} + e^{- t}}{2} + 2 \frac{e^{t} - e^{- t}}{2}} d t

= \int \frac{2 + e^{2 t} + e^{- 2 t} - 2 e^{2 t} + 2 e^{- 2 t}}{e^{t} + e^{- t} + 2 e^{t} - 2 e^{- t}} d t =_{e^{t} = u} \int \frac{2 - u^{2} + 3 u^{- 2}}{3 u - u^{- 1}} d u

= \int \frac{2 u^{2} - u^{4} + 3}{3 u^{3} - u} d u = - \int \frac{u^{4} - 2 u^{2} - 3}{3 u^{3} - u} d u l e t d e c o m p o s e

F (u) = \frac{u^{4} - 2 u^{2} - 3}{3 u^{3} - u} \Rightarrow F (u) = \frac{1}{3} \frac{3 u^{4} - 6 u^{2} - 9}{3 u^{3} - u} \Rightarrow

F (u) = \frac{1}{3} {\frac{3 (u^{3} - u) u - 3 u^{2} - 9}{3 u^{3} - u}} = \frac{1}{3} {u - \frac{3 u^{2} + 9}{3 u^{3} - u}}

= \frac{u}{3} - \frac{u^{2} + 3}{3 u^{3} - u} a n d g (u) = \frac{u^{2} + 3}{3 u^{3} - u} = \frac{u^{2} + 3}{u (3 u^{2} - 1)} = \frac{u^{2} + 3}{3 u (u - \frac{1}{\sqrt{3}}) (u + \frac{1}{\sqrt{3}})}

= \frac{a}{u} + \frac{b}{u - \frac{1}{\sqrt{3}}} + \frac{c}{u + \frac{1}{\sqrt{3}}} i t s e a z h t o f i n d a, b, c s o

\int F (u) d u = a l n ∣ u ∣ + b l n ∣ u - \frac{1}{\sqrt{3}} ∣ + c l n ∣ u + \frac{1}{\sqrt{3}} ∣ + c \Rightarrow

A = - \frac{a}{2} l n ∣ u ∣ - b l n ∣ u - \frac{1}{\sqrt{3}} ∣ - c l n ∣ u + \frac{1}{\sqrt{3}} ∣ + c

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

u = x + \sqrt{1 + x^{2}} \Rightarrow

A = - \frac{a}{2} l n ∣ x + \sqrt{1 + x^{2}} ∣ - b l n ∣ x + \sqrt{1 + x^{2}} - \frac{1}{\sqrt{3}} ∣ - c l n ∣ x + \sqrt{1 + x^{2}} + \frac{1}{\sqrt{3}} ∣ + C .

Commented by aliesam last updated on 27/May/19

Answered by ajfour last updated on 24/May/19

x = \cot θ

I = - \int \frac{cosec θ - 2 \cot θ}{cosec θ + 2 \cot θ} cosec^{2} θ d θ

= - \int \frac{1 - 2 \cos θ}{\sin^{2} θ (1 + 2 \cos θ)} d θ

= - \int \frac{2 \sin^{2} \frac{θ}{2} \frac{d θ}{2} \times 2}{8 \sin^{2} \frac{θ}{2} \cos^{4} \frac{θ}{2}}

l e t \frac{θ}{2} = ϕ

I = - \frac{1}{2} \int \sec^{2} ϕ \sec^{2} ϕ d ϕ

= - \frac{1}{2} {\sec^{2} ϕ \tan ϕ - \int 2 \sec^{2} ϕ \tan^{2} ϕ d ϕ}

= - \frac{1}{2} \sec^{2} ϕ \tan ϕ - 2 I - \tan ϕ + c

I = - \frac{1}{6} \sec^{2} \frac{θ}{2} \tan \frac{θ}{2} - \frac{1}{3} \tan \frac{θ}{2} + c

l e t \tan \frac{θ}{2} = t a n d s i n c e \tan θ = x

\frac{2 t}{1 - t^{2}} = x \Rightarrow {x t}^{2} + 2 t - x = 0

t = \frac{- 2 \pm \sqrt{4 + 4 x^{2}}}{2 x} = - \frac{1}{x} \pm \frac{\sqrt{1 + x^{2}}}{x}

I = - \frac{1}{6} (1 + t^{2}) t - \frac{t}{3} + c .

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