Question Number 152631 by mnjuly1970 last updated on 30/Aug/21
Answered by Olaf_Thorendsen last updated on 30/Aug/21
![I = ∫_0 ^1 ((ln(1−x^2 +x^4 ))/x^2 ) dx I = [−((ln(1−x^2 +x^4 ))/x)]_0 ^1 −∫_0 ^1 −(1/x)(((4x^3 −2x)/(x^4 −x^2 +1))) dx I = 2∫_0 ^1 ((2x^2 −1)/(x^4 −x^2 +1)) dx I = 2∫_0 ^1 ((2x^2 −1)/((x^2 −(√3)x+1)(x^2 +(√3)x+1))) dx etc...](https://www.tinkutara.com/question/Q152633.png)
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\left[−\frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}{{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{2}{x}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}\:{dx} \\ $$$${etc}… \\ $$
Commented by SANOGO last updated on 31/Aug/21
$${good} \\ $$
Commented by mnjuly1970 last updated on 31/Aug/21
$${thanks}\:{alot}… \\ $$
Commented by Ar Brandon last updated on 02/Sep/21
![I=2∫_0 ^1 ((2x^2 −1)/(x^4 −x^2 +1))dx= 2x^2 −1=x^2 −1+(1/2)[(x^2 −1)+(x^2 +1)] =(3/2)(x^2 −1)+(1/2)(x^2 +1) I=3∫_0 ^1 ((x^2 −1)/(x^4 −x^2 +1))dx+∫_0 ^1 ((x^2 +1)/(x^4 −x^2 +1))dx =3∫_0 ^1 ((1−(1/x^2 ))/(x^2 −1+(1/x^2 )))dx+∫_0 ^1 ((1+(1/x^2 ))/(x^2 −1+(1/x^2 )))dx =3∫_0 ^1 ((1−(1/x^2 ))/((x+(1/x))^2 −3))dx+∫_0 ^1 ((1+(1/x^2 ))/((x−(1/x))^2 +1))dx =−(√3)[argth(((x^2 +1)/( (√3)x)))]_0 ^1 +[arctan(((x^2 −1)/( (√3)x)))]_0 ^1 =((√3)/2)[ln∣((x^2 −(√3)x+1)/(x^2 +(√3)x+1))∣]_0 ^1 +(π/2)=((√3)/2)ln(((2−(√3))/(2+(√3))))+(π/2) =((√3)/2)ln((1/((2+(√3))^2 )))+(π/2)=(π/2)−(√3)ln(2+(√3)) =(1/2)(π−2(√3)ln(2+(√3)))](https://www.tinkutara.com/question/Q152700.png)
$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}={x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${I}=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\:=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{3}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:=−\sqrt{\mathrm{3}}\left[\mathrm{argth}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\mathrm{arctan}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left[\mathrm{ln}\mid\frac{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\pi}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{2}} \\ $$$$\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right)+\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\sqrt{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right) \\ $$