Question Number 53294 by gunawan last updated on 20/Jan/19
![∫_(−1/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx=...](Q53294.png)
$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \left[\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{1}/\mathrm{2}} {dx}=... \\ $$
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](Q53298.png)
$${f}\left({x}\right)=\left[\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left[\left\{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)−\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right\}^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}\leftarrow{look}\:{here}\:{f}\left(−{x}\right)=\frac{−\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}=−{f}\left({x}\right) \\ $$$${so}\:\:\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{0} \\ $$$${if}\:{we}\:{clculate}.. \\ $$$$\mathrm{2}\int_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{d}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\: \\ $$$$\mathrm{2}×\mid{ln}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\mid_{\frac{−\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{2}\left[{ln}\mid\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\mid−{ln}\mid\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\mid\right]=\mathrm{0} \\ $$