Question Number 41896 by Ugguvj3 last updated on 15/Aug/18
![∫_(−1/2) ^(1/2) [ (((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx =](https://www.tinkutara.com/question/Q41896.png)
$$\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\:\left[\:\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{1}/\mathrm{2}} {dx}\:= \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
![we have I = ∫_(−(1/2)) ^(1/2) (√((((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2))dx =∫_(−(1/2)) ^(1/2) (√((((x+1)/(x−1 ))−((x−1)/(x+1)))^2 +2((x+1)/(x−1))((x−1)/(x+1)) −2))dx =∫_(−(1/2)) ^(1/2) (√(((((x+1)^2 −(x−1)^2 )/(x^2 −1)))^2 ))dx = ∫_(−(1/2)) ^(1/2) ∣((4x)/(x^2 −1))∣ dx =4 ∫_(−(1/2)) ^(1/2) ((∣x∣)/(∣x^2 −1∣)) dx =8 ∫_0 ^(1/2) (x/(∣x^2 −1∣))dx ( fonction even) =8 ∫_0 ^(1/2) (x/(1−x^2 )) dx =−4 ∫_0 ^(1/2) ((−2x)/(1−x^2 ))dx =−4[ln∣1−x^2 ∣]_0 ^(1/2) =−4ln(1−(1/4)) =−4ln((3/(4 ))) =−4ln(3) +8ln(2) I =8ln(2)−4ln(3) .](https://www.tinkutara.com/question/Q41956.png)
$${we}\:{have}\:{I}\:=\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} \:+\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} \:−\mathrm{2}}{dx} \\ $$$$=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\sqrt{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}\:}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{2}\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\:−\mathrm{2}}{dx} \\ $$$$=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\sqrt{\left(\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)^{\mathrm{2}} }{dx}\:=\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mid\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\mid\:{dx} \\ $$$$=\mathrm{4}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mid{x}\mid}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}\:{dx}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{x}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{dx}\:\:\left(\:\:{fonction}\:{even}\right) \\ $$$$=\mathrm{8}\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\:=−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=−\mathrm{4}\left[{ln}\mid\mathrm{1}−{x}^{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\mathrm{4}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\:=−\mathrm{4}{ln}\left(\frac{\mathrm{3}}{\mathrm{4}\:}\right)\:=−\mathrm{4}{ln}\left(\mathrm{3}\right)\:+\mathrm{8}{ln}\left(\mathrm{2}\right) \\ $$$${I}\:=\mathrm{8}{ln}\left(\mathrm{2}\right)−\mathrm{4}{ln}\left(\mathrm{3}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Aug/18
![∫_((−1)/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2×(((x+1)/(x−1)))(((x−1)/(x+1)))]^(1/2) formula a^2 +b^2 −2ab=(a−b)^2 ∫_((−1)/2) ^(1/2) { (((x+1)/(x−1)))−(((x−1)/(x+1)))}dx ∫_((−1)/2) ^(1/2) (((x+1)^2 −(x−1)^2 )/(x^2 −1))dx ∫_(−(1/2)) ^(1/2) ((4x)/(x^2 −1))dx f(x)=((4x)/(x^2 −1)) f(−x)=((−4x)/(x^2 −1)) so ∫_((−1)/2) ^(1/2) ((4x)/(x^2 −1))dx=2∫_0 ^(1/2) ((4x)/(x^2 −1))dx 4∫_0 ^(1/2) ((d(x^2 −1))/(x^2 −1))dx 4∣ln∣(x^2 −1)∣∣_0 ^(1/2) 4ln(3/4)](https://www.tinkutara.com/question/Q41900.png)
$$\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left[\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}×\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${formula}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\left({a}−{b}\right)^{\mathrm{2}} \\ $$$$\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left\{\:\:\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)−\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right\}{dx} \\ $$$$\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$ \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:{f}\left(−{x}\right)=\frac{−\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${so}\:\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{d}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\mathrm{4}\mid{ln}\mid\left({x}^{\mathrm{2}} −\mathrm{1}\right)\mid\mid_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{4}{ln}\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by MJS last updated on 15/Aug/18
![(√((((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2))=4×(√(x^2 /((x^2 −1)^2 )))= =4∣(x/(x^2 −1))∣ (x/(x^2 −1)) has a simple change of sign at x=0 (and 2 changes of sign at the poles x=±1) 4∫_(−(1/2)) ^(1/2) ∣(x/(x^2 −1))∣dx=−8∫_0 ^(1/2) (x/(x^2 −1))dx=[−4ln ∣x^2 −1∣]_0 ^(1/2) = =8ln 2 −4ln 3 ≈1.15073](https://www.tinkutara.com/question/Q41908.png)
$$\sqrt{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}}=\mathrm{4}×\sqrt{\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}= \\ $$$$=\mathrm{4}\mid\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\mid \\ $$$$\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{has}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$$\left(\mathrm{and}\:\mathrm{2}\:\mathrm{changes}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{at}\:\mathrm{the}\:\mathrm{poles}\:{x}=\pm\mathrm{1}\right) \\ $$$$\mathrm{4}\underset{−\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\mid\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\mid{dx}=−\mathrm{8}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\left[−\mathrm{4ln}\:\mid{x}^{\mathrm{2}} −\mathrm{1}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$=\mathrm{8ln}\:\mathrm{2}\:−\mathrm{4ln}\:\mathrm{3}\:\approx\mathrm{1}.\mathrm{15073} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Aug/18
$${yes}\:{sir}\:{you}\:{are}\:{right}… \\ $$