Question Number 83569 by mathmax by abdo last updated on 03/Mar/20
$${calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{5}} } \\ $$
Commented by mathmax by abdo last updated on 04/Mar/20
![let A =∫_1 ^(+∞) (dx/(x^4 (3x−1)^5 )) ⇒A =∫_1 ^(+∞) (dx/(((x/(3x−1)))^4 (3x−1)^9 )) we use the changement (x/(3x−1)) =t ⇒x=3tx−t ⇒(1−3t)x=−t ⇒ x=(t/(3t−1)) and dx =((3t−1−t(3))/((3t−1)^2 ))dt =((−1)/((3t−1)^2 ))dt 3x−1 =((3t)/(3t−1))−1 =((3t−3t+1)/(3t−1)) =(1/(3t−1)) ⇒ A =∫_(1/2) ^(1/3) ((−dt)/((3t−1)^2 ×t^4 ((1/(3t−1)))^9 )) =∫_(1/3) ^(1/2) (((3t−1)^9 )/((3t−1)^2 ×t^4 ))dt =∫_(1/3) ^(1/2) (((3t−1)^7 )/t^4 )dt =∫_(1/3) ^(1/2) ((Σ_(k=0) ^7 C_7 ^k (3t)^k (−1)^(7−k) )/t^4 )dt =−∫_(1/3) ^(1/2) ((Σ_(k=0) ^7 (−3)^k C_7 ^k t^k )/t^4 )dt =−∫_(1/3) ^(1/2) Σ_(k=0) ^7 (−3)^k C_7 ^k t^(k−4) dt =−Σ_(k=0 and k≠3) ^7 (−3)^k C_7 ^k [(1/(k−3))t^(k−3) ]_(1/3) ^(1/2) −∫_(1/3) ^(1/2) (−3)^3 C_7 ^3 (dt/t) =−Σ_(k=0 and k≠3) ^7 (−3)^k (C_7 ^k /(k−3)){(1/2^(k−3) )−(1/3^(k−3) )}+27 C_7 ^3 (ln((1/2))−ln((1/3))) A=Σ_(k=0and k≠3) ^7 (((−3)^(k+1) C_7 ^k )/(k−3))((1/2^(k−3) )−(1/3^(k−3) ))+27C_7 ^3 (ln(3)−ln(2))](https://www.tinkutara.com/question/Q83650.png)
$${let}\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{5}} }\:\Rightarrow{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left(\frac{{x}}{\mathrm{3}{x}−\mathrm{1}}\right)^{\mathrm{4}} \left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{9}} } \\ $$$${we}\:{use}\:{the}\:{changement}\:\frac{{x}}{\mathrm{3}{x}−\mathrm{1}}\:={t}\:\Rightarrow{x}=\mathrm{3}{tx}−{t}\:\Rightarrow\left(\mathrm{1}−\mathrm{3}{t}\right){x}=−{t}\:\Rightarrow \\ $$$${x}=\frac{{t}}{\mathrm{3}{t}−\mathrm{1}}\:\:{and}\:{dx}\:=\frac{\mathrm{3}{t}−\mathrm{1}−{t}\left(\mathrm{3}\right)}{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\frac{−\mathrm{1}}{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\mathrm{3}{x}−\mathrm{1}\:=\frac{\mathrm{3}{t}}{\mathrm{3}{t}−\mathrm{1}}−\mathrm{1}\:=\frac{\mathrm{3}{t}−\mathrm{3}{t}+\mathrm{1}}{\mathrm{3}{t}−\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{3}{t}−\mathrm{1}}\:\Rightarrow \\ $$$${A}\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:\:\:\frac{−{dt}}{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} ×{t}^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{3}{t}−\mathrm{1}}\right)^{\mathrm{9}} }\:=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{9}} }{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} ×{t}^{\mathrm{4}} }{dt} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{7}} }{{t}^{\mathrm{4}} }{dt}\:=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:{C}_{\mathrm{7}} ^{{k}} \left(\mathrm{3}{t}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{7}−{k}} }{{t}^{\mathrm{4}} }{dt} \\ $$$$=−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:\:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{7}} ^{{k}} \:\:{t}^{{k}} }{{t}^{\mathrm{4}} }{dt} \\ $$$$=−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{7}} ^{{k}} \:{t}^{{k}−\mathrm{4}} \:{dt} \\ $$$$=−\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{7}} ^{{k}} \:\left[\frac{\mathrm{1}}{{k}−\mathrm{3}}{t}^{{k}−\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\mathrm{3}\right)^{\mathrm{3}} \:{C}_{\mathrm{7}} ^{\mathrm{3}} \:\frac{{dt}}{{t}} \\ $$$$=−\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{3}} ^{\mathrm{7}} \left(−\mathrm{3}\right)^{{k}} \:\frac{{C}_{\mathrm{7}} ^{{k}} }{{k}−\mathrm{3}}\left\{\frac{\mathrm{1}}{\mathrm{2}^{{k}−\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{{k}−\mathrm{3}} }\right\}+\mathrm{27}\:{C}_{\mathrm{7}} ^{\mathrm{3}} \:\:\left({ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$${A}=\sum_{{k}=\mathrm{0}{and}\:{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\:\:\frac{\left(−\mathrm{3}\right)^{{k}+\mathrm{1}} \:{C}_{\mathrm{7}} ^{{k}} }{{k}−\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}^{{k}−\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{{k}−\mathrm{3}} }\right)+\mathrm{27}{C}_{\mathrm{7}} ^{\mathrm{3}} \left({ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\right) \\ $$