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Question Number 95584 by turbo msup by abdo last updated on 26/May/20
calculate ∫_1 ^(+∞) (dx/(x^3 (2x+1)^4 )) 1)without use of decomposition 2)by use of decomposition
$${calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right){without}\:{use}\:{of}\:{decomposition} \\ $$$$\left.\mathrm{2}\right){by}\:{use}\:{of}\:{decomposition} \\ $$
Answered by MJS last updated on 26/May/20
Ostrogradski leads to ((960x^4 +1200x^3 +440x^2 +30x−3)/(6x^2 (2x+1)^3 ))+40∫(dx/(x(2x+1)))= =((960x^4 +1200x^3 +440x^2 +30x−3)/(6x^2 (2x+1)^3 ))+40ln (x/(2x+1)) +C ⇒ ∫_1 ^(+∞) (dx/(x^3 (2x+1)^4 ))=40ln (3/2) −((2627)/(162))
$$\mathrm{Ostrogradski}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\mathrm{960}{x}^{\mathrm{4}} +\mathrm{1200}{x}^{\mathrm{3}} +\mathrm{440}{x}^{\mathrm{2}} +\mathrm{30}{x}−\mathrm{3}}{\mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{40}\int\frac{{dx}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{960}{x}^{\mathrm{4}} +\mathrm{1200}{x}^{\mathrm{3}} +\mathrm{440}{x}^{\mathrm{2}} +\mathrm{30}{x}−\mathrm{3}}{\mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{40ln}\:\frac{{x}}{\mathrm{2}{x}+\mathrm{1}}\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{1}} {\overset{+\infty} {\int}}\:\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} }=\mathrm{40ln}\:\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{2627}}{\mathrm{162}} \\ $$
Commented by mathmax by abdo last updated on 26/May/20
thank you sir mjs
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mjs} \\ $$
Answered by mathmax by abdo last updated on 26/May/20
A =∫_1 ^(+∞) (dx/(x^3 (2x+1)^4 ))⇒A =∫_1 ^(+∞) (dx/(((x/(2x+1)))^3 (2x+1)^7 )) we do the changement (x/(2x+1)) =t ⇒ x =2tx +t ⇒(1−2t)x =t ⇒x =(t/(1−2t)) ⇒(dx/dt) =((1−2t −t(−2t))/((1−2t)^2 )) =(1/((1−2t)^2 )) and 2x+1 =((2t)/(1−2t)) +1 =(1/(1−2t)) ⇒ A = ∫_(1/3) ^(1/2) (dt/((2t−1)^2 t^3 ((1/(1−2t)))^7 )) =−∫_(1/3) ^(1/2) (((2t−1)^7 )/((2t−1)^2 t^3 ))dt =−∫_(1/3) ^(1/2) (((2t−1)^5 )/t^3 ) dt =−∫_(1/3) ^(1/2) ((Σ_(k=0) ^5 C_5 ^k (2t)^k (−1)^(5−k) )/t^3 )dt =∫_(1/3) ^(1/2) ((Σ_(k=0) ^5 2^k (−1)^k C_5 ^k t^k )/t^3 )dt =Σ_(k=0) ^5 C_5 ^k (−2)^k ∫_(1/3) ^(1/2) t^(k−3) dt =Σ_(k=0 and k≠2) ^5 (−2)^k C_5 ^k [(1/(k−2))t^(k−2) ]_(1/3) ^(1/2) +4 C_5 ^2 [lnt]_(1/3) ^(1/2) =Σ_(k=0 and k≠2) ^5 (−2)^k (C_5 ^k /(k−2)){ ((1/2))^(k−2) −((1/3))^(k−2) }+4C_5 ^2 {ln((1/2))−ln((1/3))}
$$\mathrm{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} }\Rightarrow\mathrm{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\frac{\mathrm{x}}{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{3}} \left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{7}} }\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}}{\mathrm{2x}+\mathrm{1}}\:=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{x}\:=\mathrm{2tx}\:+\mathrm{t}\:\Rightarrow\left(\mathrm{1}−\mathrm{2t}\right)\mathrm{x}\:=\mathrm{t}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{t}}{\mathrm{1}−\mathrm{2t}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}\:=\frac{\mathrm{1}−\mathrm{2t}\:−\mathrm{t}\left(−\mathrm{2t}\right)}{\left(\mathrm{1}−\mathrm{2t}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2t}\right)^{\mathrm{2}} }\:\:\:\mathrm{and}\:\mathrm{2x}+\mathrm{1}\:=\frac{\mathrm{2t}}{\mathrm{1}−\mathrm{2t}}\:+\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2t}}\:\Rightarrow \\ $$$$\mathrm{A}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{\mathrm{dt}}{\left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2t}}\right)^{\mathrm{7}} }\:=−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{3}} }\mathrm{dt} \\ $$$$=−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{5}} }{\mathrm{t}^{\mathrm{3}} }\:\mathrm{dt}\:\:=−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \left(\mathrm{2t}\right)^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{5}−\mathrm{k}} }{\mathrm{t}^{\mathrm{3}} }\mathrm{dt} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{2}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \:\:\mathrm{t}^{\mathrm{k}} }{\mathrm{t}^{\mathrm{3}} }\mathrm{dt} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \left(−\mathrm{2}\right)^{\mathrm{k}} \:\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{t}^{\mathrm{k}−\mathrm{3}} \:\mathrm{dt} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{2}} ^{\mathrm{5}} \:\left(−\mathrm{2}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{2}}\mathrm{t}^{\mathrm{k}−\mathrm{2}} \right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:+\mathrm{4}\:\mathrm{C}_{\mathrm{5}} ^{\mathrm{2}} \:\left[\mathrm{lnt}\right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{2}} ^{\mathrm{5}} \:\left(−\mathrm{2}\right)^{\mathrm{k}} \:\frac{\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{2}}\left\{\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}−\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{k}−\mathrm{2}} \right\}+\mathrm{4C}_{\mathrm{5}} ^{\mathrm{2}} \left\{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right\} \\ $$
Answered by mathmax by abdo last updated on 27/May/20
thank you sir mjs 2) let decompose F(x) =(1/(x^3 (2x+1)^4 )) ⇒ 2^4 F(x) =(1/(x^3 (x+(1/2))^4 )) =Σ_(i=1) ^3 (a_i /x^i ) +Σ_(i=1) ^4 (b_i /((x+(1/2))^i )) a_i ? we find D_2 (0)for f(x) =(x+(1/2))^(−4) f(x)=f(0) +(x/(1!))f^′ (0) +(x^2 /(2!))f^((2)) (0) +(x^3 /(3!))ξ(x) f(0) =(1/2^(−4) ) ,f^′ (x) =−4(x+(1/2))^(−5) ⇒f^′ (0) =((−4)/2^(−5) ) f^((2)) (x) =20(x+(1/2))^(−6) ⇒f^((2)) (0) =((20)/2^(−6) ) ⇒ f(x) =2^4 −4 .2^5 x +10 .2^6 x^2 +(x^3 /(3!))ξ(x) ⇒ ((f(x))/x^3 ) =(2^4 /x^3 ) −((4.2^5 )/x^2 ) +((10.2^6 )/x) +(1/(3!))ξ(x) ⇒a_1 =10.2^6 a_2 =−4.2^5 and a_3 =2^4 let find b_i we determine D_3 (−(1/2)) for g(x) =x^(−3) g(x) =g(−(1/2))+((x+(1/2))/(1!))g^′ (−(1/2)) +(((x+(1/2))^2 )/(2!))g^((2)) (−(1/2)) +(((x+(1/2))^3 )/(3!))g^((3)) (−(1/2))+(((x+(1/2))^4 )/(4!))ξ(x) g(−(1/2))=(−(1/2))^(−3) ,g^′ (x)=−3x^(−4) ⇒g^′ (−(1/2))=−3(−(1/2))^(−4) g^((2)) (x) =12 x^(−5) ⇒g^((2)) (−(1/2)) =12(−(1/2))^(−5) g^((3)) (x) =−60 x^(−6) ⇒g^((3)) (−(1/2))=−60(−(1/2))^(−6) ⇒g(x)=(−(1/2))^(−3) −3(−(1/2))^(−4) (x+(1/2))+6(−(1/2))^(−5) (x+(1/2))^2 −10 (−(1/2))^(−6) (x+(1/2))^3 +(((x+(1/2))^4 )/(4!))ξ(x) =−2^3 −3 .2^4 (x+(1/2))−6 .2^5 (x+(1/2))^2 −10.2^6 (x+(1/2))^3 +(((x+(1/2))^4 )/(4!))ξ(x) ⇒ ((g(x))/((x+(1/2))^4 )) =−(2^3 /((x+(1/2))^4 ))−((3.2^4 )/((x+(1/2))^3 ))−((6.2^5 )/((x+(1/2))^2 ))−((10.2^6 )/((x+(1/2)))) +..⇒ ⇒b_1 =−10.2^6 ,b_2 =−6.2^5 , b_3 =−3.2^4 , b_4 =−2^3 ⇒ 2^4 F(x) =((10.2^6 )/x) −((4.2^5 )/x^2 ) +(2^4 /x^3 ) −((10.2^6 )/((x+(1/2)))) −((6.2^5 )/((x+(1/2))^2 ))−((3.2^4 )/((x+(1/2))^3 )) −(2^3 /((x+(1/2))^4 )) ⇒ F(x) =((10.2^2 )/x)−((4.2)/x^2 ) +(1/x^3 ) −((10.2^2 )/((x+(1/2))))−((6.2)/((x+(1/2))^2 ))−(3/((x+(1/2))^3 )) −(2^(−1) /((x+(1/2))^4 )) now its eazy to find ∫_1 ^(+∞) F(x)dx...
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mjs} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{2}^{\mathrm{4}} \mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }\:=\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{3}} \:\frac{\mathrm{a}_{\mathrm{i}} }{\mathrm{x}^{\mathrm{i}} }\:+\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{\mathrm{b}_{\mathrm{i}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{i}} } \\ $$$$\mathrm{a}_{\mathrm{i}} ?\:\:\mathrm{we}\:\mathrm{find}\:\mathrm{D}_{\mathrm{2}} \left(\mathrm{0}\right)\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{4}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{0}\right)\:+\frac{\mathrm{x}}{\mathrm{1}!}\mathrm{f}^{'} \left(\mathrm{0}\right)\:+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}\xi\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{−\mathrm{4}} }\:\:,\mathrm{f}^{'} \left(\mathrm{x}\right)\:=−\mathrm{4}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{5}} \:\Rightarrow\mathrm{f}^{'} \left(\mathrm{0}\right)\:=\frac{−\mathrm{4}}{\mathrm{2}^{−\mathrm{5}} } \\ $$$$\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{20}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{6}} \:\Rightarrow\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{20}}{\mathrm{2}^{−\mathrm{6}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{2}^{\mathrm{4}} \:−\mathrm{4}\:.\mathrm{2}^{\mathrm{5}} \:\mathrm{x}\:+\mathrm{10}\:.\mathrm{2}^{\mathrm{6}} \:\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}\xi\left(\mathrm{x}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{3}} }\:=\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{3}} }\:−\frac{\mathrm{4}.\mathrm{2}^{\mathrm{5}} }{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{10}.\mathrm{2}^{\mathrm{6}} }{\mathrm{x}}\:+\frac{\mathrm{1}}{\mathrm{3}!}\xi\left(\mathrm{x}\right)\:\Rightarrow\mathrm{a}_{\mathrm{1}} =\mathrm{10}.\mathrm{2}^{\mathrm{6}} \\ $$$$\mathrm{a}_{\mathrm{2}} =−\mathrm{4}.\mathrm{2}^{\mathrm{5}} \:\:\mathrm{and}\:\mathrm{a}_{\mathrm{3}} =\mathrm{2}^{\mathrm{4}} \:\mathrm{let}\:\mathrm{find}\:\mathrm{b}_{\mathrm{i}} \\ $$$$\mathrm{we}\:\mathrm{determine}\:\mathrm{D}_{\mathrm{3}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{for}\:\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{x}^{−\mathrm{3}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{g}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}!}\mathrm{g}^{'} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}!}\mathrm{g}^{\left(\mathrm{2}\right)} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$+\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}!}\mathrm{g}^{\left(\mathrm{3}\right)} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}!}\xi\left(\mathrm{x}\right) \\ $$$$\mathrm{g}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{3}} \:,\mathrm{g}^{'} \left(\mathrm{x}\right)=−\mathrm{3x}^{−\mathrm{4}} \:\Rightarrow\mathrm{g}^{'} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{4}} \\ $$$$\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{12}\:\mathrm{x}^{−\mathrm{5}} \:\Rightarrow\mathrm{g}^{\left(\mathrm{2}\right)} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{12}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{5}} \\ $$$$\mathrm{g}^{\left(\mathrm{3}\right)} \left(\mathrm{x}\right)\:=−\mathrm{60}\:\mathrm{x}^{−\mathrm{6}} \:\Rightarrow\mathrm{g}^{\left(\mathrm{3}\right)} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{60}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{6}} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{3}} −\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{4}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{6}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{5}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$−\mathrm{10}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{6}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \:+\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}!}\xi\left(\mathrm{x}\right) \\ $$$$=−\mathrm{2}^{\mathrm{3}} \:−\mathrm{3}\:.\mathrm{2}^{\mathrm{4}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{6}\:.\mathrm{2}^{\mathrm{5}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\mathrm{10}.\mathrm{2}^{\mathrm{6}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$+\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}!}\xi\left(\mathrm{x}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{g}\left(\mathrm{x}\right)}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }\:=−\frac{\mathrm{2}^{\mathrm{3}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }−\frac{\mathrm{3}.\mathrm{2}^{\mathrm{4}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\frac{\mathrm{6}.\mathrm{2}^{\mathrm{5}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{\mathrm{10}.\mathrm{2}^{\mathrm{6}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:+..\Rightarrow \\ $$$$\Rightarrow\mathrm{b}_{\mathrm{1}} =−\mathrm{10}.\mathrm{2}^{\mathrm{6}} \:,\mathrm{b}_{\mathrm{2}} =−\mathrm{6}.\mathrm{2}^{\mathrm{5}} \:,\:\mathrm{b}_{\mathrm{3}} =−\mathrm{3}.\mathrm{2}^{\mathrm{4}} \:,\:\mathrm{b}_{\mathrm{4}} =−\mathrm{2}^{\mathrm{3}} \:\Rightarrow \\ $$$$\mathrm{2}^{\mathrm{4}} \:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{10}.\mathrm{2}^{\mathrm{6}} }{\mathrm{x}}\:−\frac{\mathrm{4}.\mathrm{2}^{\mathrm{5}} }{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{3}} }\:−\frac{\mathrm{10}.\mathrm{2}^{\mathrm{6}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:−\frac{\mathrm{6}.\mathrm{2}^{\mathrm{5}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{\mathrm{3}.\mathrm{2}^{\mathrm{4}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$−\frac{\mathrm{2}^{\mathrm{3}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{10}.\mathrm{2}^{\mathrm{2}} }{\mathrm{x}}−\frac{\mathrm{4}.\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:−\frac{\mathrm{10}.\mathrm{2}^{\mathrm{2}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}−\frac{\mathrm{6}.\mathrm{2}}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$−\frac{\mathrm{2}^{−\mathrm{1}} }{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }\:\:\mathrm{now}\:\mathrm{its}\:\mathrm{eazy}\:\mathrm{to}\:\mathrm{find}\:\int_{\mathrm{1}} ^{+\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}… \\ $$

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