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Question Number 201110 by emilagazade last updated on 29/Nov/23

∫(1/( (√((x−a)^3 ))+(√((x+a)^3 ))))dx

$$\int\frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)^{\mathrm{3}} }+\sqrt{\left({x}+{a}\right)^{\mathrm{3}} }}{dx} \\ $$

Answered by Frix last updated on 29/Nov/23

(√p)+(√q)=(√(p+2(√(pq))+q)) ∫(dx/( (x−a)^(3/2) +(x+a)^(3/2) ))= =(1/( (√2)))∫(dx/((x(x^2 +3a^2 )+(x^2 −a^2 )^(3/2) )^(1/3) ))= (use t=((x+(√(x^2 −a^2 )))/a) ⇔ x=((a(t^2 +1))/(2t)) → dx=((a(t^2 −1))/(2t^2 ))dt) =(1/( (√(2a))))∫((t^2 −1)/(t^(3/2) (t^2 +3)))dt =^(u=(√t)) ((√2)/( (√a)))∫((u^4 −1)/(u^2 (u^4 +3)))du= (decompose...) =((√2)/(3(√a)u))+(2/( 3^(5/4) (√a)))(tan^(−1) ((((√2)u)/3^(1/4) )+1) +tan^(−1) ((((√2)u)/3^(1/4) )−1))+(1/(3^(5/4) (√a)))ln ((u^2 −3^(1/4) (√2)u+(√3))/(u^2 +3^(1/4) (√2)u+(√3))) Now insert u=((√(x+(√(x^2 −a^2 ))))/( (√a)))

$$\sqrt{{p}}+\sqrt{{q}}=\sqrt{{p}+\mathrm{2}\sqrt{{pq}}+{q}} \\ $$$$\int\frac{{dx}}{\:\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\left({x}\left({x}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} }= \\ $$$$\left(\mathrm{use}\:{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\:\Leftrightarrow\:{x}=\frac{{a}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{{a}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} }{dt}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({t}^{\mathrm{2}} +\mathrm{3}\right)}{dt}\:\overset{{u}=\sqrt{{t}}} {=}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{a}}}\int\frac{{u}^{\mathrm{4}} −\mathrm{1}}{{u}^{\mathrm{2}} \left({u}^{\mathrm{4}} +\mathrm{3}\right)}{du}= \\ $$$$\left(\mathrm{decompose}...\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{{a}}{u}}+\frac{\mathrm{2}}{\:\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{4}}} \sqrt{{a}}}\left(\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\sqrt{\mathrm{2}}{u}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} }+\mathrm{1}\right)\:+\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\sqrt{\mathrm{2}}{u}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} }−\mathrm{1}\right)\right)+\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{4}}} \sqrt{{a}}}\mathrm{ln}\:\frac{{u}^{\mathrm{2}} −\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\mathrm{2}}{u}+\sqrt{\mathrm{3}}}{{u}^{\mathrm{2}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\mathrm{2}}{u}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Now}\:\mathrm{insert} \\ $$$${u}=\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}}{\:\sqrt{{a}}} \\ $$

Commented by emilagazade last updated on 29/Nov/23

thank you a lot Sir

$${thank}\:{you}\:{a}\:{lot}\:{Sir} \\ $$

Commented by Frix last updated on 29/Nov/23

I also tried (1/((x−a)^(3/2) +(x+a)^(3/2) ))=(((x+a)^(3/2) −(x−a)^(3/2) )/(2a(3x^2 +a^2 ))) leading to (1/(2a))(∫(((x+a)^(3/2) )/(3x^2 +a^2 ))dx−∫(((x−a)^(3/2) )/(3x^2 +a^2 ))dx) but it′s not better...

$$\mathrm{I}\:\mathrm{also}\:\mathrm{tried} \\ $$$$\frac{\mathrm{1}}{\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}{a}\left(\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{leading}\:\mathrm{to} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\int\frac{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}−\int\frac{\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{better}... \\ $$

Commented by emilagazade last updated on 29/Nov/23

this one I tried too. but the book suggests to do substitution which leads to partial fractions.

$${this}\:{one}\:{I}\:{tried}\:{too}.\:{but}\:{the}\:{book}\:{suggests}\:{to}\:{do}\:{substitution}\:{which}\:{leads}\:{to}\:{partial}\:{fractions}. \\ $$

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