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Question Number 36801 by rahul 19 last updated on 05/Jun/18

∫ ((1+x^4 )/((1−x^4 )^(3/2) )) dx = A ∫ A = B Find B ? Assume integration of constant=0.

$$\int\:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx}\:=\:{A}\: \\ $$$$\int\:\mathrm{A}\:=\:\mathrm{B} \\ $$$$\mathrm{Find}\:\mathrm{B}\:? \\ $$$$\mathrm{Assume}\:\mathrm{integration}\:\mathrm{of}\:\mathrm{constant}=\mathrm{0}. \\ $$

Answered by MJS last updated on 05/Jun/18

A=(x/(√(1−x^4 ))) B=(1/2)arcsin x^2 A: ∫((1+x^4 )/((1−x^4 )^(3/2) ))dx looks like it′s ((p(x))/(√(1−x^4 ))) with p(x) is a polynome let′s try (d/dx)(((p(x))/(√(1−x^4 ))))=(d/dx)(p(x)(1−x^4 )^(−(1/2)) )= =p′(x)(1−x^4 )^(−(1/2)) +p(x)(2x^3 (1−x^4 )^(−(3/2)) )= =((p′(x)(1−x^4 )+2p(x)x^3 )/((1−x^4 )^(3/2) )) ⇒ p′(x)(1−x^4 )+2p(x)x^3 =1+x^4 let′s try p(x)=ax+b; p′(x)=a a+2bx^3 +ax^4 =1+x^4 a=1; b=0 p(x)=x ∫((1+x^4 )/((1−x^4 )^(3/2) ))dx=(x/(√(1−x^4 ))) B: ∫(x/(√(1−x^4 )))dx= [t=x^2 → dx=(dt/(2x))] =(1/2)∫(dt/(√(1−t^2 )))=(1/2)arcsin t= =(1/2)arcsin x^2

$${A}=\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:{x}^{\mathrm{2}} \\ $$$$ \\ $$$${A}: \\ $$$$\int\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$$$\mathrm{looks}\:\mathrm{like}\:\mathrm{it}'\mathrm{s}\:\frac{{p}\left({x}\right)}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\mathrm{with}\:{p}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynome} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{p}\left({x}\right)}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\right)=\frac{{d}}{{dx}}\left({p}\left({x}\right)\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)= \\ $$$$={p}'\left({x}\right)\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} +{p}\left({x}\right)\left(\mathrm{2}{x}^{\mathrm{3}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \right)= \\ $$$$=\frac{{p}'\left({x}\right)\left(\mathrm{1}−{x}^{\mathrm{4}} \right)+\mathrm{2}{p}\left({x}\right){x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\Rightarrow\:{p}'\left({x}\right)\left(\mathrm{1}−{x}^{\mathrm{4}} \right)+\mathrm{2}{p}\left({x}\right){x}^{\mathrm{3}} =\mathrm{1}+{x}^{\mathrm{4}} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:{p}\left({x}\right)={ax}+{b};\:{p}'\left({x}\right)={a} \\ $$$${a}+\mathrm{2}{bx}^{\mathrm{3}} +{ax}^{\mathrm{4}} =\mathrm{1}+{x}^{\mathrm{4}} \\ $$$${a}=\mathrm{1};\:{b}=\mathrm{0} \\ $$$${p}\left({x}\right)={x} \\ $$$$\int\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$$$ \\ $$$${B}: \\ $$$$\int\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:{t}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:{x}^{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

∫((1+x^4 )/((1−x^4 )(√(1−x^4 )) ))dx x^2 =(1/t) x=t^((−1)/2) dx=((−1)/2)×(1/t^(3/2) )dt =((−1)/2)∫(dt/t^(3/2) )×((1+(1/t^2 ))/(1−(1/t^2 )))×(1/(√(1−(1/t^2 )))) =((−1)/2)∫(dt/t^(3/2) )×((t^2 +1)/(t^2 −1))×(t/(√(t^2 −1))) =((−1)/2)∫((t^2 +1)/(t^2 −1))×(1/(√t))×(1/((√t) ((√(t−(1/t)))))dt =((−1)/2)∫((1+(1/t^2 ))/(1−(1/t^2 )))×(1/t)×(dt/(√(t−(1/t)))) ((−1)/2)∫((1+(1/t^2 ))/(t−(1/t)))×(dt/(√(t−(1/t)))) =((−1)/2)∫(dk/k^(3/2) ) when(k=t−(1/t)) =((−1)/2)×(k^((−1)/2) /((−1)/2))+c =(1/(√k))+c =(1/(√(t−(1/t))))+c =(1/(√((1/x^2 )−x^2 )))+c so A=(1/(√((1/x^2 )−x^2 ))) ∫Adx=B ∫(x/(√(1−x^4 )))dx y=x^2 dy=2xdx ∫(dy/(2×(√(1−y^2 )))) =(1/2)sin^(−1) (y) =(1/2)sin^(−1) (x^2 ) is B

$$ \\ $$$$\int\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }\:}{dx} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{{t}}\:\:{x}={t}^{\frac{−\mathrm{1}}{\mathrm{2}}} \:\:{dx}=\frac{−\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }×\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }×\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}×\frac{{t}}{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}×\frac{\mathrm{1}}{\sqrt{{t}}}×\frac{\mathrm{1}}{\sqrt{{t}}\:\left(\sqrt{{t}−\frac{\mathrm{1}}{{t}}}\right.}{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}×\frac{\mathrm{1}}{{t}}×\frac{{dt}}{\sqrt{{t}−\frac{\mathrm{1}}{{t}}}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}−\frac{\mathrm{1}}{{t}}}×\frac{{dt}}{\sqrt{{t}−\frac{\mathrm{1}}{{t}}}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\:\:\:\:{when}\left({k}={t}−\frac{\mathrm{1}}{{t}}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}×\frac{{k}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\frac{−\mathrm{1}}{\mathrm{2}}}+{c} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{k}}}+{c} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{t}−\frac{\mathrm{1}}{{t}}}}+{c} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} }}+{c} \\ $$$$ \\ $$$${so}\:{A}=\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} }} \\ $$$$\int{Adx}={B} \\ $$$$\int\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{dx}\: \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${dy}=\mathrm{2}{xdx} \\ $$$$\int\frac{{dy}}{\mathrm{2}×\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left({y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)\:\:{is}\:{B} \\ $$$$ \\ $$

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