∫_(−1/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx=...


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Question Number 53294 by gunawan last updated on 20/Jan/19

$\int_{- 1 / 2}^{1 / 2} {[{(\frac{x + 1}{x - 1})}^{2} + {(\frac{x - 1}{x + 1})}^{2} - 2]}^{1 / 2} d x = . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
	$f(x)=[(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) =[{(((x+1)/(x−1)))−(((x−1)/(x+1)))}^2 ]^(1/2) =(((x+1)^2 −(x−1)^2 )/((x^2 −1))) =((4x)/(x^2 −1)) ∫_((−1)/2) ^(1/2) ((4x)/(x^2 −1))dx←look here f(−x)=((−4x)/(x^2 −1))=−f(x) so ∫_((−1)/2) ^(1/2) ((4x)/(x^2 −1))=0 if we clculate.. 2∫_((−1)/2) ^(1/2) ((d(x^2 −1))/(x^2 −1)) 2×∣ln(x^2 −1)∣_((−1)/2) ^(1/2) 2[ln∣((1/4)−1)∣−ln∣((1/4)−1)∣]=0$
	$f (x) = {[{(\frac{x + 1}{x - 1})}^{2} + {(\frac{x - 1}{x + 1})}^{2} - 2]}^{\frac{1}{2}}$ $= {[{(\frac{x + 1}{x - 1}) - (\frac{x - 1}{x + 1})}^{2}]}^{\frac{1}{2}}$ $= \frac{{(x + 1)}^{2} - {(x - 1)}^{2}}{(x^{2} - 1)}$ $= \frac{4 x}{x^{2} - 1}$ $\int_{\frac{- 1}{2}}^{\frac{1}{2}} \frac{4 x}{x^{2} - 1} d x \leftarrow l o o k h e r e f (- x) = \frac{- 4 x}{x^{2} - 1} = - f (x)$ $s o \int_{\frac{- 1}{2}}^{\frac{1}{2}} \frac{4 x}{x^{2} - 1} = 0$ $i f w e c l c u l a t e . .$ $2 \int_{\frac{- 1}{2}}^{\frac{1}{2}} \frac{d (x^{2} - 1)}{x^{2} - 1}$ $2 \times ∣ l n (x^{2} - 1) ∣_{\frac{- 1}{2}}^{\frac{1}{2}}$ $2 [l n ∣ (\frac{1}{4} - 1) ∣ - l n ∣ (\frac{1}{4} - 1) ∣] = 0$
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