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Question Number 84894 by Power last updated on 17/Mar/20

Commented by abdomathmax last updated on 18/Mar/20

I=∫(√((ax+b)/(cx+d)))dx we do the changement(√((ax+b)/(cx+d)))=t ⇒ ((ax+b)/(cx+d))=t^2 ⇒ax+b=ct^2 x +dt^2 ⇒(a−ct^2 )x=dt^2 −b ⇒ x=((dt^2 −b)/(a−ct^2 )) ⇒dx =(((2dt)(a−ct^2 )−(dt^2 −b)(−2ct))/((a−ct^2 )^2 ))dt =((2adt−2dct^3 +2cdt^3 −2bct)/((ct^2 −a)^2 )) =((2(ad−bc)t)/((ct^2 −a)^2 )) ⇒ I=∫ t×((2(ad−bc)t)/((ct^2 −a)^2 ))dt =2(ad−bc)∫ (t^2 /((ct^2 −a)^2 ))dt =((2(ad−bc))/c)∫ ((ct^2 −a+a)/((ct^2 −a)^2 ))dt =((2(ad−bc))/c) ∫ (dt/(ct^2 −a)) +((2a(ad−bc))/c)∫ (dt/((ct^2 −a)^2 )) we have ∫ (dt/(ct^2 −a)) =∫ (dt/(((√c)t)^2 −((√a))^2 )) =∫ (dt/(((√c)t−(√a))((√c)t+(√a)))) =∫ ((1/((√c)t−(√a)))−(1/((√c)t+(√a))))dt =(1/(2(√(ac))))ln∣(((√c)t−(√a))/((√c)t +(√a)))∣ +c ∫ (dt/((ct^2 −a)^2 )) =∫ (1/(((√c)t−(√a))^2 ((√c)t+(√a))^2 )) let decompose F(t)=(1/(((√c)t−(√a))^2 ((√c)t+(√a))^2 )) F(t)=(α/((√c)t −(√a))) +(β/(((√c)t−(√a))^2 )) +(λ/((√c)t+(√a))) +(ρ/(((√c)t+(√a))^2 )) ...be continued...

$${I}=\int\sqrt{\frac{{ax}+{b}}{{cx}+{d}}}{dx}\:{we}\:{do}\:{the}\:{changement}\sqrt{\frac{{ax}+{b}}{{cx}+{d}}}={t}\:\Rightarrow \\ $$$$\frac{{ax}+{b}}{{cx}+{d}}={t}^{\mathrm{2}} \:\Rightarrow{ax}+{b}={ct}^{\mathrm{2}} {x}\:+{dt}^{\mathrm{2}} \:\Rightarrow\left({a}−{ct}^{\mathrm{2}} \right){x}={dt}^{\mathrm{2}} −{b}\:\Rightarrow \\ $$$${x}=\frac{{dt}^{\mathrm{2}} −{b}}{{a}−{ct}^{\mathrm{2}} }\:\:\Rightarrow{dx}\:=\frac{\left(\mathrm{2}{dt}\right)\left({a}−{ct}^{\mathrm{2}} \right)−\left({dt}^{\mathrm{2}} −{b}\right)\left(−\mathrm{2}{ct}\right)}{\left({a}−{ct}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}{adt}−\mathrm{2}{dct}^{\mathrm{3}} +\mathrm{2}{cdt}^{\mathrm{3}} −\mathrm{2}{bct}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\left({ad}−{bc}\right){t}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}=\int\:\:{t}×\frac{\mathrm{2}\left({ad}−{bc}\right){t}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }{dt}\:=\mathrm{2}\left({ad}−{bc}\right)\int\:\:\frac{{t}^{\mathrm{2}} }{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}\left({ad}−{bc}\right)}{{c}}\int\:\:\frac{{ct}^{\mathrm{2}} −{a}+{a}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}\left({ad}−{bc}\right)}{{c}}\:\int\:\:\frac{{dt}}{{ct}^{\mathrm{2}} −{a}}\:+\frac{\mathrm{2}{a}\left({ad}−{bc}\right)}{{c}}\int\:\:\frac{{dt}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\int\:\:\frac{{dt}}{{ct}^{\mathrm{2}} −{a}}\:=\int\:\:\frac{{dt}}{\left(\sqrt{{c}}{t}\right)^{\mathrm{2}} −\left(\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{dt}}{\left(\sqrt{{c}}{t}−\sqrt{{a}}\right)\left(\sqrt{{c}}{t}+\sqrt{{a}}\right)} \\ $$$$=\int\:\left(\frac{\mathrm{1}}{\sqrt{{c}}{t}−\sqrt{{a}}}−\frac{\mathrm{1}}{\sqrt{{c}}{t}+\sqrt{{a}}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{ac}}}{ln}\mid\frac{\sqrt{{c}}{t}−\sqrt{{a}}}{\sqrt{{c}}{t}\:+\sqrt{{a}}}\mid\:+{c} \\ $$$$\int\:\:\:\frac{{dt}}{\left({ct}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }\:=\int\:\frac{\mathrm{1}}{\left(\sqrt{{c}}{t}−\sqrt{{a}}\right)^{\mathrm{2}} \left(\sqrt{{c}}{t}+\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{1}}{\left(\sqrt{{c}}{t}−\sqrt{{a}}\right)^{\mathrm{2}} \left(\sqrt{{c}}{t}+\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$${F}\left({t}\right)=\frac{\alpha}{\sqrt{{c}}{t}\:−\sqrt{{a}}}\:+\frac{\beta}{\left(\sqrt{{c}}{t}−\sqrt{{a}}\right)^{\mathrm{2}} }\:+\frac{\lambda}{\sqrt{{c}}{t}+\sqrt{{a}}}\:+\frac{\rho}{\left(\sqrt{{c}}{t}+\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$$...{be}\:{continued}... \\ $$

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