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Question Number 112251 by bobhans last updated on 07/Sep/20
∫ (dx/((x^4 −1)(√(x^2 +1))))
$$\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Answered by john santu last updated on 07/Sep/20
∫ (dx/((x^4 −1)(√(x^2 +1)))) ? setting: tan r = x ⇒dx = sec^2 r dr I=∫ (dx/((x^2 −1)(x^2 +1)(√(x^2 +1)))) I= ∫ ((sec^2 r dr)/((tan^2 r−1)sec^3 r)) I= ∫ ((cos r dr)/(tan^2 r−1))=∫((cos^3 r dr)/(sin^2 r−cos^2 r)) I= ∫ (((1−sin^2 r)cos r dr)/(2sin^2 r−1)) let sin r = q ⇒I=∫ ((1−q^2 )/(2q^2 −1)) dq I= (1/2)∫ ((1/(2q^2 −1))−1) dq I= (1/4)∫(−2+(1/(q(√2)−1))−(1/(q(√2)+1)))dq I= (1/4)(−2q+(1/( (√2)))ln ∣q(√2) −1∣−(1/( (√2)))ln ∣q(√2) +1∣ )+ c I=(1/4)(((−2x)/( (√(x^2 +1))))+(1/( (√2)))ln ∣((x(√2) −(√(x^2 +1)))/(x(√2) +(√(x^2 +1))))∣ ) + c
$$\int\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:? \\ $$$$\:{setting}:\:\mathrm{tan}\:{r}\:=\:{x}\:\Rightarrow{dx}\:=\:\mathrm{sec}\:^{\mathrm{2}} {r}\:{dr} \\ $$$${I}=\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${I}=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {r}\:{dr}}{\left(\mathrm{tan}\:^{\mathrm{2}} {r}−\mathrm{1}\right)\mathrm{sec}\:^{\mathrm{3}} {r}} \\ $$$${I}=\:\int\:\frac{\mathrm{cos}\:{r}\:{dr}}{\mathrm{tan}\:^{\mathrm{2}} {r}−\mathrm{1}}=\int\frac{\mathrm{cos}\:^{\mathrm{3}} {r}\:{dr}}{\mathrm{sin}\:^{\mathrm{2}} {r}−\mathrm{cos}\:^{\mathrm{2}} {r}} \\ $$$${I}=\:\int\:\frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {r}\right)\mathrm{cos}\:{r}\:{dr}}{\mathrm{2sin}\:^{\mathrm{2}} {r}−\mathrm{1}} \\ $$$${let}\:\mathrm{sin}\:{r}\:=\:{q}\:\Rightarrow{I}=\int\:\frac{\mathrm{1}−{q}^{\mathrm{2}} }{\mathrm{2}{q}^{\mathrm{2}} −\mathrm{1}}\:{dq} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\frac{\mathrm{1}}{\mathrm{2}{q}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}\right)\:{dq} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}\int\left(−\mathrm{2}+\frac{\mathrm{1}}{{q}\sqrt{\mathrm{2}}−\mathrm{1}}−\frac{\mathrm{1}}{{q}\sqrt{\mathrm{2}}+\mathrm{1}}\right){dq} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{2}{q}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid{q}\sqrt{\mathrm{2}}\:−\mathrm{1}\mid−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid{q}\sqrt{\mathrm{2}}\:+\mathrm{1}\mid\:\right)+\:{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{−\mathrm{2}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{x}\sqrt{\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}\sqrt{\mathrm{2}}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\mid\:\right)\:+\:{c} \\ $$
Commented by bemath last updated on 07/Sep/20
santuyyy
$${santuyyy} \\ $$

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