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Question Number 123452 by Eric002 last updated on 25/Nov/20
∫(dx/( (x^2 +n)(√(x^2 +a))))
$$\int\frac{{dx}}{\:\left({x}^{\mathrm{2}} +{n}\right)\sqrt{{x}^{\mathrm{2}} +{a}}} \\ $$
Commented by MJS_new last updated on 25/Nov/20
depends on the values of a and n and their relation...
$$\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{and}\:{n}\:\mathrm{and}\:\mathrm{their} \\ $$$$\mathrm{relation}… \\ $$
Answered by TANMAY PANACEA last updated on 25/Nov/20
t^2 =((x^2 +a)/(x^2 +n)) t^2 x^2 +t^2 n=x^2 +a x^2 (t^2 −1)=a−t^2 n x^2 =((a−t^2 n)/(t^2 −1)) 2xdx=(((t^2 −1)(−2tn)−(a−t^2 n)(2t))/((t^2 −1)^2 ))dt 2xdx=((−2t^3 n+2tn−2at+2t^3 n)/((t^2 −1)^2 ))dt=((2t(n−a))/((t^2 −1)^2 ))dt xdx=((t(n−a))/((t^2 −1)^2 ))dt dx=((t(n−a))/((t^2 −1)^2 ))×(1/x)=((t(n−a))/((t^2 −1)^2 ))×((√(t^2 −1))/( (√(a−t^2 n ))))dt x^2 +n =((a−t^2 n)/(t^2 −1))+n=((a−t^2 n+nt^2 −n)/(t^2 −1))=((a−n)/(t^2 −1)) x^2 +a=((a−t^2 n)/(t^2 −1))+a=((a−t^2 n+at^2 −a)/(t^2 −1))=((t^2 (a−n))/(t^2 −1)) ∫(dx/((x^2 +n)(√(x^2 +a)))) ∫((t(a−n))/((t^2 −1)^2 ))×(√(t^2 −1)) ×(1/( (√(a−t^2 n))))×((t^2 −1)/(a−n))×(1/( (√((t^2 (a−n))/(t^2 −1)))))×dt ∫((t(a−n))/((t^2 −1)^2 ))×(t^2 −1)^2 ×(1/( (√(a−t^2 n)) ))×(1/((a−n)^(3/2) ))×(dt/t) ∫(1/((a−n)^(1/2) ))×(dt/( (√n) ((√((a/n)−t^2 )) ))) (1/((an−n^2 )^(1/2) ))∫(dt/( (√((a/n)−t^2 )))) (1/((an−n^2 )^(1/2) ))sin^(−1) ((t/( (√(a/n)))))+C (1/((an−n^2 )^(1/2) ))×sin^(−1) (((√((x^2 +a)/(x^2 +n)))/( (√(a/n)))))+C pls chk
$${t}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} +{a}}{{x}^{\mathrm{2}} +{n}} \\ $$$${t}^{\mathrm{2}} {x}^{\mathrm{2}} +{t}^{\mathrm{2}} {n}={x}^{\mathrm{2}} +{a} \\ $$$${x}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)={a}−{t}^{\mathrm{2}} {n} \\ $$$${x}^{\mathrm{2}} =\frac{{a}−{t}^{\mathrm{2}} {n}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{2}{xdx}=\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left(−\mathrm{2}{tn}\right)−\left({a}−{t}^{\mathrm{2}} {n}\right)\left(\mathrm{2}{t}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}{xdx}=\frac{−\mathrm{2}{t}^{\mathrm{3}} {n}+\mathrm{2}{tn}−\mathrm{2}{at}+\mathrm{2}{t}^{\mathrm{3}} {n}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{2}{t}\left({n}−{a}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${xdx}=\frac{{t}\left({n}−{a}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${dx}=\frac{{t}\left({n}−{a}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{{x}}=\frac{{t}\left({n}−{a}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }×\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{{a}−{t}^{\mathrm{2}} {n}\:}}{dt} \\ $$$${x}^{\mathrm{2}} +{n} \\ $$$$=\frac{{a}−{t}^{\mathrm{2}} {n}}{{t}^{\mathrm{2}} −\mathrm{1}}+{n}=\frac{{a}−{t}^{\mathrm{2}} {n}+{nt}^{\mathrm{2}} −{n}}{{t}^{\mathrm{2}} −\mathrm{1}}=\frac{{a}−{n}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${x}^{\mathrm{2}} +{a}=\frac{{a}−{t}^{\mathrm{2}} {n}}{{t}^{\mathrm{2}} −\mathrm{1}}+{a}=\frac{{a}−{t}^{\mathrm{2}} {n}+{at}^{\mathrm{2}} −{a}}{{t}^{\mathrm{2}} −\mathrm{1}}=\frac{{t}^{\mathrm{2}} \left({a}−{n}\right)}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{n}\right)\sqrt{{x}^{\mathrm{2}} +{a}}} \\ $$$$\int\frac{{t}\left({a}−{n}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }×\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:×\frac{\mathrm{1}}{\:\sqrt{{a}−{t}^{\mathrm{2}} {n}}}×\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{a}−{n}}×\frac{\mathrm{1}}{\:\sqrt{\frac{{t}^{\mathrm{2}} \left({a}−{n}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}}}×{dt} \\ $$$$\int\frac{{t}\left({a}−{n}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }×\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\:\sqrt{{a}−{t}^{\mathrm{2}} {n}}\:}×\frac{\mathrm{1}}{\left({a}−{n}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }×\frac{{dt}}{{t}} \\ $$$$\int\frac{\mathrm{1}}{\left({a}−{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }×\frac{{dt}}{\:\sqrt{{n}}\:\left(\sqrt{\frac{{a}}{{n}}−{t}^{\mathrm{2}} }\:\right)} \\ $$$$\frac{\mathrm{1}}{\left({an}−{n}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int\frac{{dt}}{\:\sqrt{\frac{{a}}{{n}}−{t}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left({an}−{n}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{sin}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\frac{{a}}{{n}}}}\right)+{C} \\ $$$$\frac{\mathrm{1}}{\left({an}−{n}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }×{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\frac{{x}^{\mathrm{2}} +{a}}{{x}^{\mathrm{2}} +{n}}}}{\:\sqrt{\frac{{a}}{{n}}}}\right)+{C} \\ $$$${pls}\:{chk} \\ $$
Commented by TANMAY PANACEA last updated on 25/Nov/20
pld chk error if sny
$${pld}\:{chk}\:{error}\:{if}\:{sny} \\ $$
Commented by MJS_new last updated on 25/Nov/20
what if an−n^2 ≤0 or (a/n)<0?
$$\mathrm{what}\:\mathrm{if}\:{an}−{n}^{\mathrm{2}} \leqslant\mathrm{0}\:\mathrm{or}\:\frac{{a}}{{n}}<\mathrm{0}? \\ $$
Answered by nueron last updated on 25/Nov/20
Commented by MJS_new last updated on 25/Nov/20
yeah. but for a=−4∧n=3 you would probably choose a different path...
$$\mathrm{yeah}.\:\mathrm{but}\:\mathrm{for}\:{a}=−\mathrm{4}\wedge{n}=\mathrm{3}\:\mathrm{you}\:\mathrm{would} \\ $$$$\mathrm{probably}\:\mathrm{choose}\:\mathrm{a}\:\mathrm{different}\:\mathrm{path}… \\ $$
Answered by TANMAY PANACEA last updated on 25/Nov/20
x=(√a) tanθ dx=(√a) sec^2 θdθ ∫(((√a) sec^2 θ dθ)/((atan^2 θ+n)(√a) secθ)) ∫(dθ/(cosθ(a((sin^2 θ)/(cos^2 θ))+n))) ∫((cosθ dθ)/(asin^2 θ+n(1−sin^2 θ))) ∫((d(sinθ))/(n+(a−n)sin^2 θ)) (1/((a−n)))∫((d(sinθ))/( ((√(n/(a−n))) )^2 +sin^2 θ)) (1/((a−n)))×((√(a−n))/( (√n)))tan^(−1) (((sinθ)/( (√(n/(a−n))))))+C (1/( (√n) ×(√(a−n))))×tan^(−1) (((x/( (√(a+x^2 ))))/( (√(n/(a−n))))))+C
$${x}=\sqrt{{a}}\:{tan}\theta \\ $$$${dx}=\sqrt{{a}}\:{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int\frac{\sqrt{{a}}\:{sec}^{\mathrm{2}} \theta\:{d}\theta}{\left({atan}^{\mathrm{2}} \theta+{n}\right)\sqrt{{a}}\:{sec}\theta}\: \\ $$$$\int\frac{{d}\theta}{{cos}\theta\left({a}\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}+{n}\right)} \\ $$$$\int\frac{{cos}\theta\:{d}\theta}{{asin}^{\mathrm{2}} \theta+{n}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right)} \\ $$$$\int\frac{{d}\left({sin}\theta\right)}{{n}+\left({a}−{n}\right){sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\left({a}−{n}\right)}\int\frac{{d}\left({sin}\theta\right)}{\:\left(\sqrt{\frac{{n}}{{a}−{n}}}\:\:\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\left({a}−{n}\right)}×\frac{\sqrt{{a}−{n}}}{\:\sqrt{{n}}}{tan}^{−\mathrm{1}} \left(\frac{{sin}\theta}{\:\sqrt{\frac{{n}}{{a}−{n}}}}\right)+{C} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{n}}\:×\sqrt{{a}−{n}}}×{tan}^{−\mathrm{1}} \left(\frac{\frac{{x}}{\:\sqrt{{a}+{x}^{\mathrm{2}} }}}{\:\sqrt{\frac{{n}}{{a}−{n}}}}\right)+{C} \\ $$$$ \\ $$
Answered by Bird last updated on 25/Nov/20
I =∫ (dx/((x^2 +n)(√(x^2 +a)))) a>0 we do the changement x=(√a)sh(t) ⇒ I=∫ (((√a)ch(t))/( (√a)ch(t)(ash^2 t +n)))dt =∫ (dt/(ash^2 t +n)) =∫ (dt/(a.((ch(2t)−1)/2)+n)) =∫ ((2dt)/(ach(2t)−a+2n)) =∫ ((2dt)/(a((e^(2t) +e^(−2t) )/2)+2n−a)) =∫ ((4dt)/(a(e^(2t) +e^(−2t) )+4n−2a)) =_(e^(2t) =u) 4 ∫ (du/(2u{au+au^(−1) +4n−2a})) =2 ∫ (du/(au^2 +a+(4n−2a)u)) =2∫ (du/(au^(2 ) +(4n−2a)u+a)) Δ^′ =(2n−a)^2 −a^2 =4n^2 −4na =4n(n−a) if n<a ⇒Δ^′ <0 ⇒ au^2 +(4n−2a)u +a =a{u^2 +2(((2n−a))/a))u +(((2n−a)^2 )/a^2 ) +a−(((2n−a)^2 )/a^2 )} =a{(u+((2n−a)/a))^2 +((a^3 −(2n−a)^2 )/a^2 )} in this case we do the changement u+((2n−a)/a)=(√((a^3 −(2n−a)^2 )/a^2 ))z) if n>a Δ^′ >0 ⇒u_1 =((a−2n+2(√(n^2 −na)))/a) u_2 =((a−2n−2(√(n^2 −na)))/a) ⇒I =(2/a)∫ (du/((u−u_1 )(u−u_2 ))) =(2/(a(u_1 −u_2 )))∫ ((1/(u−u_1 ))−(1/(u−u_2 )))du =(2/(a(u_1 −u_2 )))ln∣((u−u_1 )/(u−u_2 ))∣ +c u=e^(2t) snd t=argsh((x/( (√a)))) =ln((x/( (√a)))+(√(1+(x^2 /a)))) =ln(((x+(√(a+x^2 )))/( (√a)))) ⇒ u=(((x+(√(a+x^2 )))/( (√a))))^2 ....
$${I}\:=\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{n}\right)\sqrt{{x}^{\mathrm{2}} +{a}}} \\ $$$${a}>\mathrm{0}\:{we}\:{do}\:{the}\:{changement} \\ $$$${x}=\sqrt{{a}}{sh}\left({t}\right)\:\Rightarrow \\ $$$${I}=\int\:\:\frac{\sqrt{{a}}{ch}\left({t}\right)}{\:\sqrt{{a}}{ch}\left({t}\right)\left({ash}^{\mathrm{2}} {t}\:+{n}\right)}{dt} \\ $$$$=\int\:\:\:\frac{{dt}}{{ash}^{\mathrm{2}} {t}\:+{n}}\:=\int\:\frac{{dt}}{{a}.\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}+{n}} \\ $$$$=\int\:\:\frac{\mathrm{2}{dt}}{{ach}\left(\mathrm{2}{t}\right)−{a}+\mathrm{2}{n}} \\ $$$$=\int\:\:\frac{\mathrm{2}{dt}}{{a}\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}+\mathrm{2}{n}−{a}} \\ $$$$=\int\:\frac{\mathrm{4}{dt}}{{a}\left({e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \right)+\mathrm{4}{n}−\mathrm{2}{a}} \\ $$$$=_{{e}^{\mathrm{2}{t}} \:={u}} \:\:\:\:\mathrm{4}\:\int\:\:\:\frac{{du}}{\mathrm{2}{u}\left\{{au}+{au}^{−\mathrm{1}} +\mathrm{4}{n}−\mathrm{2}{a}\right\}} \\ $$$$=\mathrm{2}\:\int\:\:\:\frac{{du}}{{au}^{\mathrm{2}} +{a}+\left(\mathrm{4}{n}−\mathrm{2}{a}\right){u}} \\ $$$$=\mathrm{2}\int\:\:\frac{{du}}{{au}^{\mathrm{2}\:} +\left(\mathrm{4}{n}−\mathrm{2}{a}\right){u}+{a}} \\ $$$$\Delta^{'} \:=\left(\mathrm{2}{n}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$=\mathrm{4}{n}^{\mathrm{2}} −\mathrm{4}{na}\:=\mathrm{4}{n}\left({n}−{a}\right) \\ $$$${if}\:{n}<{a}\:\:\Rightarrow\Delta^{'} <\mathrm{0}\:\Rightarrow \\ $$$${au}^{\mathrm{2}} \:+\left(\mathrm{4}{n}−\mathrm{2}{a}\right){u}\:+{a} \\ $$$$={a}\left\{{u}^{\mathrm{2}} \:+\mathrm{2}\left(\frac{\left.\mathrm{2}{n}−{a}\right)}{{a}}\right){u}\:+\frac{\left(\mathrm{2}{n}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right. \\ $$$$\left.+{a}−\frac{\left(\mathrm{2}{n}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right\} \\ $$$$={a}\left\{\left({u}+\frac{\mathrm{2}{n}−{a}}{{a}}\right)^{\mathrm{2}} \:+\frac{{a}^{\mathrm{3}} −\left(\mathrm{2}{n}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right\} \\ $$$${in}\:{this}\:{case}\:{we}\:{do}\:{the}\:{changement} \\ $$$$\left.{u}+\frac{\mathrm{2}{n}−{a}}{{a}}=\sqrt{\frac{{a}^{\mathrm{3}} −\left(\mathrm{2}{n}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{z}\right) \\ $$$${if}\:{n}>{a}\:\:\Delta^{'} \:>\mathrm{0}\:\Rightarrow{u}_{\mathrm{1}} =\frac{{a}−\mathrm{2}{n}+\mathrm{2}\sqrt{{n}^{\mathrm{2}} −{na}}}{{a}} \\ $$$${u}_{\mathrm{2}} =\frac{{a}−\mathrm{2}{n}−\mathrm{2}\sqrt{{n}^{\mathrm{2}} −{na}}}{{a}} \\ $$$$\Rightarrow{I}\:=\frac{\mathrm{2}}{{a}}\int\:\:\frac{{du}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{{a}\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)}\int\:\:\left(\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right){du} \\ $$$$=\frac{\mathrm{2}}{{a}\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)}{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\:+{c} \\ $$$${u}={e}^{\mathrm{2}{t}} \:{snd}\:{t}={argsh}\left(\frac{{x}}{\:\sqrt{{a}}}\right)\: \\ $$$$={ln}\left(\frac{{x}}{\:\sqrt{{a}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}\right) \\ $$$$={ln}\left(\frac{{x}+\sqrt{{a}+{x}^{\mathrm{2}} }}{\:\sqrt{{a}}}\right)\:\Rightarrow \\ $$$${u}=\left(\frac{{x}+\sqrt{{a}+{x}^{\mathrm{2}} }}{\:\sqrt{{a}}}\right)^{\mathrm{2}} …. \\ $$
Commented by Bird last updated on 25/Nov/20
if a<0 we do the changement x=(√(−a))ch(t)
$${if}\:{a}<\mathrm{0}\:{we}\:{do}\:{the}\:{changement} \\ $$$${x}=\sqrt{−{a}}{ch}\left({t}\right) \\ $$

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