Question Number 143142 by Ar Brandon last updated on 10/Jun/21
$$\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 10/Jun/21
![F(x) = ∫((x^2 −1)/(x^2 +1)).(1/( (√(1+x^4 )))) dx F(x) = ∫((x−(1/x))/(x+(1/x))).(1/( sign(x).x(√(x^2 +(1/x^2 ))))) dx F(x) = sign(x)∫(1/(x+(1/x))).(1/( (√((x+(1/x))^2 −2))))(1−(1/x^2 ))dx Let u = x+(1/x) F(u) = sign(x)∫(1/u).(1/( (√(u^2 −2)))) du Let u = (√2)cht F(t) = sign(x)∫(1/( (√2)cht)).(1/( (√2)sht)) (√2)sht dt F(t) = sign(x)(1/( (√2)))∫(1/( cht)) dt F(t) = sign(x)(1/( (√2)))∫(2/( e^t +e^(−t) )) dt F(t) = sign(x)(√2)∫(e^t /( 1+e^(2t) )) dt F(t) = sign(x)(√2)arctan(e^t ) F(u) = sign(x)(√2)arctan(e^(argch((u/( (√2))))) ) F(u) = sign(x)(√2)arctan(e^(ln((u/( (√2)))+(√((u/2)−1)))) ) F(u) = sign(x)(√2)arctan((u/( (√2)))+(√((u/2)−1))) F(x) = sign(x)(√2)arctan(((x+(1/x))/( (√2)))+(√(((x+(1/x))/2)−1))) F(x) = sign(x)(√2)arctan[(1/( (√2)x))(x^2 +1+(√(1−x^2 )))]](https://www.tinkutara.com/question/Q143149.png)
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{x}−\frac{\mathrm{1}}{{x}}}{{x}+\frac{\mathrm{1}}{{x}}}.\frac{\mathrm{1}}{\:\mathrm{sign}\left({x}\right).{x}\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{sign}\left({x}\right)\int\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}}.\frac{\mathrm{1}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}\: \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{sign}\left({x}\right)\int\frac{\mathrm{1}}{{u}}.\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{2}}}\:{du} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{\mathrm{2}}\mathrm{ch}{t} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\mathrm{sign}\left({x}\right)\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{ch}{t}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{sh}{t}}\:\sqrt{\mathrm{2}}\mathrm{sh}{t}\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\mathrm{sign}\left({x}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\:\mathrm{ch}{t}}\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\mathrm{sign}\left({x}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{2}}{\:{e}^{{t}} +{e}^{−{t}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\int\frac{{e}^{{t}} }{\:\mathrm{1}+{e}^{\mathrm{2}{t}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left({e}^{{t}} \right) \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left({e}^{\mathrm{argch}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)} \right) \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left({e}^{\mathrm{ln}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}}{\mathrm{2}}−\mathrm{1}}\right)} \right) \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}}{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{x}+\frac{\mathrm{1}}{{x}}}{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{sign}\left({x}\right)\sqrt{\mathrm{2}}\mathrm{arctan}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{x}}\left({x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\right] \\ $$