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Question Number 113628 by mathmax by abdo last updated on 14/Sep/20
find ∫  (dx/((x+1)(√(x^2 −1))+(x−1)(√(x^2  +1))))
$$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}+\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$
Answered by MJS_new last updated on 16/Sep/20
∫(dx/((x+1)(√(x^2 −1))+(x−1)(√(x^2 +1))))=  =∫(((x+1)(√(x^2 −1))−(x−1)(√(x^2 +1)))/(2(x−1)(2x^2 +x+1)))dx=  =(1/4)∫((√(x^2 −1))/(x−1))dx−(1/2)∫((√(x^2 +1))/(2x^2 +x+1))dx−(1/4)∫(((2x+1)(√(x^2 −1)))/(2x^2 +x+1))dx    (1/4)∫((√(x^2 −1))/(x−1))dx=     ( )       [t=x+(√(x^2 −1)) → dx=((√(x^2 −1))/(x+(√(x^2 −1))))dt]  =(1/8)∫(((t+1)^2 )/t^2 )dt=((t^2 −1)/(8t))+(1/4)ln t =  =(1/4)(√(x^2 −1))+(1/4)ln (x+(√(x^2 −1)))    −(1/2)∫((√(x^2 +1))/(2x^2 +x+1))dx=       [t=x+(√(x^2 +1)) → dx=((√(x^2 +1))/(x+(√(x^2 +1))))]  =−(1/4)∫(((t^2 +1)^2 )/(t(t^4 +t^3 −t+1)))dt  now we must decompose  t^4 +t^3 −t+1=(t^2 +αt+β)(t^2 +γt+δ)  α=((1+(√(5+4(√2))))/2)  β=((1+(√2))/2)+(((√(5+4(√2)))+(√(−35+28(√2))))/8)  γ=((1−(√(5+4(√2))))/2)  δ=((1+(√2))/2)−(((√(5+4(√2)))+(√(−35+28(√2))))/8)  I′m too tired now    −(1/4)∫(((2x+1)(√(x^2 −1)))/(2x^2 +x+1))dx=       [t=x+(√(x^2 −1)) → dx=((√(x^2 −1))/(x+(√(x^2 −1))))]  =−(1/8)∫(((t^2 −1)^2 (t^2 +t+1))/(t^2 (t^4 +t^3 +4t^2 +t+1)))dt  again to decompose  t^4 +t^3 +4t^2 +t+1=(t^2 +at+b)(t^2 +ct+d)  a=((1+(√(−11+8(√2))))/2)  b=((1+2(√2))/2)+(((√(−11+8(√2)))+(√(77+56(√2))))/8)  c=((1−(√(−11+8(√2))))/2)  d=((1+2(√2))/2)−(((√(−11+8(√2)))+(√(77+56(√2))))/8)  please continue if you need it, I go to bed...
$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$=\int\frac{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}−\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}−\mathrm{1}}{dx}=\:\:\:\:\:\left(\:\right) \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }{dt}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8}{t}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{t}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$ \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}\left({t}^{\mathrm{4}} +{t}^{\mathrm{3}} −{t}+\mathrm{1}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{decompose} \\ $$$${t}^{\mathrm{4}} +{t}^{\mathrm{3}} −{t}+\mathrm{1}=\left({t}^{\mathrm{2}} +\alpha{t}+\beta\right)\left({t}^{\mathrm{2}} +\gamma{t}+\delta\right) \\ $$$$\alpha=\frac{\mathrm{1}+\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}+\sqrt{−\mathrm{35}+\mathrm{28}\sqrt{\mathrm{2}}}}{\mathrm{8}} \\ $$$$\gamma=\frac{\mathrm{1}−\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\delta=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}+\sqrt{−\mathrm{35}+\mathrm{28}\sqrt{\mathrm{2}}}}{\mathrm{8}} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{tired}\:\mathrm{now} \\ $$$$ \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} \left({t}^{\mathrm{4}} +{t}^{\mathrm{3}} +\mathrm{4}{t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{dt} \\ $$$$\mathrm{again}\:\mathrm{to}\:\mathrm{decompose} \\ $$$${t}^{\mathrm{4}} +{t}^{\mathrm{3}} +\mathrm{4}{t}^{\mathrm{2}} +{t}+\mathrm{1}=\left({t}^{\mathrm{2}} +\mathfrak{a}{t}+\mathfrak{b}\right)\left({t}^{\mathrm{2}} +\mathfrak{c}{t}+\mathfrak{d}\right) \\ $$$$\mathfrak{a}=\frac{\mathrm{1}+\sqrt{−\mathrm{11}+\mathrm{8}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathfrak{b}=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{−\mathrm{11}+\mathrm{8}\sqrt{\mathrm{2}}}+\sqrt{\mathrm{77}+\mathrm{56}\sqrt{\mathrm{2}}}}{\mathrm{8}} \\ $$$$\mathfrak{c}=\frac{\mathrm{1}−\sqrt{−\mathrm{11}+\mathrm{8}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathfrak{d}=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{−\mathrm{11}+\mathrm{8}\sqrt{\mathrm{2}}}+\sqrt{\mathrm{77}+\mathrm{56}\sqrt{\mathrm{2}}}}{\mathrm{8}} \\ $$$$\mathrm{please}\:\mathrm{continue}\:\mathrm{if}\:\mathrm{you}\:\mathrm{need}\:\mathrm{it},\:\mathrm{I}\:\mathrm{go}\:\mathrm{to}\:\mathrm{bed}… \\ $$
Commented by Dwaipayan Shikari last updated on 16/Sep/20
(1/4)∫(√((x+1)/(x−1)))  (1/4)∫((x+1)/( (√(x^2 −1))))=(1/8)∫((2x)/( (√(x^2 −1))))+(1/4)∫(1/( (√(x^2 −1))))=(1/4)(√(x^2 −1))+(1/4)log(x+(√(x^2 −1)))
$$\frac{\mathrm{1}}{\mathrm{4}}\int\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{2}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}}{log}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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