Calculate-I-1-x-1-x-1-x-dx-Indication-poser-t-1-x-1-x-

Question Number 167882 by LEKOUMA last updated on 28/Mar/22

C a l c u l a t e

I = \int \frac{1}{x} (\sqrt{\frac{1 - x}{1 + x}}) d x

I n d i c a t i o n p o s e r t = \sqrt{\frac{1 - x}{1 + x}}

Answered by ArielVyny last updated on 29/Mar/22

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

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

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

I = - \int \frac{1 + t^{2}}{1 - t^{2}} . t . \frac{t}{1 + t^{2}} d t = \int \frac{- 1 - t^{2} + 1}{1 - t^{2}} d t

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

t = s i n u \to d t = c o s u d u

I = t - \int \frac{1}{{c o s}^{2} u} c o s u d u = \sqrt{\frac{1 - x}{1 + x}} + \int \frac{1}{c o s u} d u

I = \sqrt{\frac{1 - x}{1 + x}} + \int \frac{c o s u}{(1 - {s i n}^{2} u)} d u

= \dots . . - \frac{1}{2} (- \int - \frac{c o s u}{1 - s i n u} + \int \frac{c o s u}{1 + s i n u} d u)

= \dots . . - \frac{1}{2} [- l n (1 - s i n u) + l n (1 + s i n u)]

= \dots . . - \frac{1}{2} l n (\frac{1 + s i n u}{1 - s i n u})

= \dots . . - \frac{1}{2} l n (\frac{1 + t}{1 - t}) = \sqrt{\frac{1 - x}{1 + x} -} \frac{1}{2} l n (\frac{1 + \sqrt{\frac{1 - x}{1 + x}}}{1 - \sqrt{\frac{1 - x}{1 + x}}})

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

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

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

= \sqrt{\frac{1 - x}{1 + x}} - \frac{1}{2} l n [\frac{1 + \sqrt{1 - x^{2}}}{x}]

Commented by peter frank last updated on 28/Mar/22

thanks

Commented by MJS_new last updated on 29/Mar/22

this is wrong . what you do in the 1^{st} and 3^{rd}

lines makes no sense .

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

d x = \frac{d t}{\frac{d}{d x} [\frac{\sqrt{1 - x}}{\sqrt{1 + x}}]} = \frac{d t}{- \frac{1}{\sqrt{1 - x} \sqrt{{(1 + x)}^{3}}}} = - \sqrt{1 - x} \sqrt{{(1 + x)}^{3}} d t

\Rightarrow

\int \frac{\sqrt{1 - x}}{x \sqrt{1 + x}} d x = \int \frac{\sqrt{1 - x}}{x \sqrt{1 + x}} \times (- \sqrt{1 - x} \sqrt{{(1 + x)}^{3}} d t) =

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

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

Commented by ArielVyny last updated on 29/Mar/22

y e s y o u h a v e r i g h t . i h a v e f o r g o t t n e s q u a r e

Answered by MJS_new last updated on 28/Mar/22

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

[t = \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \to d x = - \sqrt{1 - x} \sqrt{{(1 + x)}^{3}} d t]

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

= 2 \arctan t + \ln \frac{t - 1}{t + 1} =

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

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