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Question Number 63822 by mathmax by abdo last updated on 09/Jul/19

find ∫ (x^2 +1)(√((x+1)/(x−2)))dx

$${find}\:\int\:\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}}{dx} \\ $$

Commented by mathmax by abdo last updated on 11/Jul/19

let A =∫ (x^2 +1)(√((x+1)/(x−2)))dx changement (√((x+1)/(x−2)))=t give ((x+1)/(x−2)) =t^(2 ) ⇒((x−2 +3)/(x−2)) =t^2 ⇒1+(3/(x−2)) =t^2 ⇒(3/(x−2)) =t^2 −1 ⇒ ((x−2)/3) =(1/(t^2 −1)) ⇒x−2 =(3/(t^2 −1)) ⇒dx =((−6t)/((t^2 −1)^2 ))dt ⇒ A = ∫ {((3/(t^2 −1)) +2)^2 +1}t ((−6t)/((t^2 −1)^2 ))dt =∫ ((−6t^2 )/((t^2 −1)^2 )){ (((2t^2 +1)/(t^2 −1)))^(2 ) +1}dt =∫ ((−6t^2 )/((t^2 −1)^2 ))(((2t^2 +1)^2 +(t^2 −1)^2 )/((t^2 −1)^2 ))dt =∫ ((−6t^2 { 4t^4 +4t^2 +1+t^4 −2t^2 +1})/((t^2 −1)^4 ))dt =−6 ∫ ((t^2 {5t^4 +2t^2 +2})/((t^2 −1)^4 )) dt =−6 ∫ ((5t^6 +2t^4 +2t^2 )/((t^2 −1)^4 ))dt let decompose F(t)=((5t^6 +2t^4 +2t^2 )/((t^2 −1)^4 )) =(((...))/((t−1)^4 (t+1)^4 )) =Σ_(i=1) ^4 (a_i /((t−1)^i )) +Σ_(i=1) ^4 (b_i /((t+1)^i )) changement t−1 =u give F(t)=G(u) =((5(u+1)^6 +2(u+1)^4 +2(u+1)^2 )/(u^4 (u+2)^4 )) let find D_3 (0) for f(u) =(1/((u+2)^4 )) =(u+2)^(−4) ⇒f(u) =f(0)+uf^′ (0) +(u^2 /2)f^((2)) (0) +(u^3 /(3!))f^((3)) (0) +(u^3 /(3!))ξ(u) f(0) =2^(−4) ,f^′ (u)=−4(u+2)^(−5) ⇒f^′ (0) =−4.2^(−5) f^((2)) (u) =20(u+2)^(−6) ⇒f^((2)) (0) =20.2^(−6) f^((3)) (u) =−120(u+2)^(−7) ⇒f^((3)) (0) =−120 .2^(−7) ⇒ f(u) =2^(−4) −4.2^(−5) u +10. 2^(−6) u^2 −20. 2^(−7) u^3 +(u^3 /(3!))ξ(u) ⇒ G(u) =((5(u+1)^6 +2(u+1)^4 +2(u+1)^2 )/u^4 ){2^(−4) −4.2^(−5) u +10.2^(−6) u^2 −20.2^(−7) u^3 +(u^3 /(3!))ξ(u)}.....be continued....

$${let}\:{A}\:=\int\:\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}}{dx}\:{changement}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}}={t}\:{give} \\ $$$$\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}\:={t}^{\mathrm{2}\:} \:\Rightarrow\frac{{x}−\mathrm{2}\:+\mathrm{3}}{{x}−\mathrm{2}}\:={t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+\frac{\mathrm{3}}{{x}−\mathrm{2}}\:={t}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{3}}{{x}−\mathrm{2}}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{3}}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{x}−\mathrm{2}\:=\frac{\mathrm{3}}{{t}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{dx}\:=\frac{−\mathrm{6}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$${A}\:=\:\int\:\left\{\left(\frac{\mathrm{3}}{{t}^{\mathrm{2}} −\mathrm{1}}\:+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}{t}\:\:\:\frac{−\mathrm{6}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int\:\:\:\frac{−\mathrm{6}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\:\left(\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}\:} +\mathrm{1}\right\}{dt} \\ $$$$=\int\:\:\frac{−\mathrm{6}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\frac{\left(\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int\:\:\frac{−\mathrm{6}{t}^{\mathrm{2}} \left\{\:\:\mathrm{4}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}+{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right\}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt} \\ $$$$=−\mathrm{6}\:\int\:\:\:\frac{{t}^{\mathrm{2}} \left\{\mathrm{5}{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}\right\}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }\:{dt}\:=−\mathrm{6}\:\int\:\:\frac{\mathrm{5}{t}^{\mathrm{6}} \:+\mathrm{2}{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{5}{t}^{\mathrm{6}} \:+\mathrm{2}{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{\left(...\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} \left({t}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{{a}_{{i}} }{\left({t}−\mathrm{1}\right)^{{i}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{{b}_{{i}} }{\left({t}+\mathrm{1}\right)^{{i}} }\:{changement}\:{t}−\mathrm{1}\:={u}\:{give} \\ $$$${F}\left({t}\right)={G}\left({u}\right)\:=\frac{\mathrm{5}\left({u}+\mathrm{1}\right)^{\mathrm{6}} \:+\mathrm{2}\left({u}+\mathrm{1}\right)^{\mathrm{4}} \:+\mathrm{2}\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{4}} \left({u}+\mathrm{2}\right)^{\mathrm{4}} }\:{let}\:{find}\:{D}_{\mathrm{3}} \left(\mathrm{0}\right)\:{for} \\ $$$${f}\left({u}\right)\:=\frac{\mathrm{1}}{\left({u}+\mathrm{2}\right)^{\mathrm{4}} }\:=\left({u}+\mathrm{2}\right)^{−\mathrm{4}} \:\Rightarrow{f}\left({u}\right)\:={f}\left(\mathrm{0}\right)+{uf}^{'} \left(\mathrm{0}\right)\:+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right) \\ $$$$+\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}\xi\left({u}\right) \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{2}^{−\mathrm{4}} \:\:\:\:,{f}^{'} \left({u}\right)=−\mathrm{4}\left({u}+\mathrm{2}\right)^{−\mathrm{5}} \:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=−\mathrm{4}.\mathrm{2}^{−\mathrm{5}} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({u}\right)\:=\mathrm{20}\left({u}+\mathrm{2}\right)^{−\mathrm{6}} \:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{20}.\mathrm{2}^{−\mathrm{6}} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({u}\right)\:=−\mathrm{120}\left({u}+\mathrm{2}\right)^{−\mathrm{7}} \:\Rightarrow{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−\mathrm{120}\:.\mathrm{2}^{−\mathrm{7}} \:\Rightarrow \\ $$$${f}\left({u}\right)\:=\mathrm{2}^{−\mathrm{4}} \:−\mathrm{4}.\mathrm{2}^{−\mathrm{5}} {u}\:+\mathrm{10}.\:\mathrm{2}^{−\mathrm{6}} {u}^{\mathrm{2}} \:−\mathrm{20}.\:\mathrm{2}^{−\mathrm{7}} \:{u}^{\mathrm{3}} \:\:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}\xi\left({u}\right)\:\Rightarrow \\ $$$${G}\left({u}\right)\:=\frac{\mathrm{5}\left({u}+\mathrm{1}\right)^{\mathrm{6}} \:+\mathrm{2}\left({u}+\mathrm{1}\right)^{\mathrm{4}} \:+\mathrm{2}\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{4}} }\left\{\mathrm{2}^{−\mathrm{4}} −\mathrm{4}.\mathrm{2}^{−\mathrm{5}} {u}\:+\mathrm{10}.\mathrm{2}^{−\mathrm{6}} {u}^{\mathrm{2}} \right. \\ $$$$\left.−\mathrm{20}.\mathrm{2}^{−\mathrm{7}} \:{u}^{\mathrm{3}} \:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}\xi\left({u}\right)\right\}.....{be}\:{continued}.... \\ $$$$ \\ $$$$ \\ $$

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