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Question Number 101269 by mathmax by abdo last updated on 01/Jul/20
calculate ∫_4 ^(+∞)      (dx/((x−2)^5 (x+3)^7 ))
$$\mathrm{calculate}\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{5}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{7}} } \\ $$
Answered by mathmax by abdo last updated on 05/Jul/20
I =∫_4 ^(∞ )   (dx/((x−2)^5 (x+3)^7 )) ⇒ I =∫_4 ^∞  (dx/((((x+3)/(x−2)))^7 (x−2)^(12) )) we do the cha7gement  ((x+3)/(x−2)) =t ⇒x+3 =tx−2t ⇒(1−t)x =−2t−3 ⇒x =((2t+3)/(t−1)) ⇒  (dx/dt) =((2(t−1)−(2t+3))/((t−1)^2 )) =((−5)/((t−1)^2 )) and x−2 =((2t+3)/(t−1))−2 =((2t+3−2t+2)/(t−1)) =(5/(t−1)) ⇒  I =∫_(7/2) ^1  ((−5dt)/((t−1)^2 .t^7 .((5/(t−1)))^(12) )) =(1/5^(11) ) ∫_1 ^(7/2)  (((t−1)^(12) )/((t−1)^2  t^7 ))dt  ⇒  5^(11) ×I =∫_1 ^(7/2)  (((t−1)^(10) )/t^7 )dt =∫_1 ^(7/2)  ((Σ_(k=0) ^(10 )  C_(10) ^k  t^k (−1)^(10−k) )/t^7 )dt  =Σ_(k=0) ^(10) (−1)^(k )  C_(10) ^k   ∫_1 ^(7/(2 ))  t^(k−7)  dt  =Σ_(k=0 and k≠6) ^(10)  (−1)^k  C_(10) ^k  [(1/(k−6))t^(k−6) ]_1 ^(7/2)   +C_(10) ^6  [ln∣t∣]_1 ^(7/2)   =Σ_(k=0 and k≠6) ^(10)  (−1)^k  (C_(10) ^k /(k−6)){((7/2))^(k−6) −1} +C_(10) ^6  ln((7/2)) ⇒  I =(1/5^(11) ) Σ_(k=0 and k≠6) ^(10)  (((−1)^k  C_(10) ^k )/(k−6)){ ((7/2))^(k−6) −1}+(C_(10) ^6 /5^(11) )ln((7/2)) .
$$\mathrm{I}\:=\int_{\mathrm{4}} ^{\infty\:} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{5}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{7}} }\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{4}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{\mathrm{7}} \left({x}−\mathrm{2}\right)^{\mathrm{12}} }\:{we}\:{do}\:{the}\:{cha}\mathrm{7}{gement} \\ $$$$\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\:={t}\:\Rightarrow\mathrm{x}+\mathrm{3}\:=\mathrm{tx}−\mathrm{2t}\:\Rightarrow\left(\mathrm{1}−\mathrm{t}\right)\mathrm{x}\:=−\mathrm{2t}−\mathrm{3}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{2t}+\mathrm{3}}{\mathrm{t}−\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}\:=\frac{\mathrm{2}\left(\mathrm{t}−\mathrm{1}\right)−\left(\mathrm{2t}+\mathrm{3}\right)}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{5}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{and}\:\mathrm{x}−\mathrm{2}\:=\frac{\mathrm{2t}+\mathrm{3}}{\mathrm{t}−\mathrm{1}}−\mathrm{2}\:=\frac{\mathrm{2t}+\mathrm{3}−\mathrm{2t}+\mathrm{2}}{\mathrm{t}−\mathrm{1}}\:=\frac{\mathrm{5}}{\mathrm{t}−\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\frac{\mathrm{7}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{−\mathrm{5dt}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} .\mathrm{t}^{\mathrm{7}} .\left(\frac{\mathrm{5}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{12}} }\:=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{11}} }\:\int_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{12}} }{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{7}} }\mathrm{dt}\:\:\Rightarrow \\ $$$$\mathrm{5}^{\mathrm{11}} ×\mathrm{I}\:=\int_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{10}} }{\mathrm{t}^{\mathrm{7}} }\mathrm{dt}\:=\int_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{10}\:} \:\mathrm{C}_{\mathrm{10}} ^{\mathrm{k}} \:\mathrm{t}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{10}−\mathrm{k}} }{\mathrm{t}^{\mathrm{7}} }\mathrm{dt} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{10}} \left(−\mathrm{1}\right)^{\mathrm{k}\:} \:\mathrm{C}_{\mathrm{10}} ^{\mathrm{k}} \:\:\int_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}\:}} \:\mathrm{t}^{\mathrm{k}−\mathrm{7}} \:\mathrm{dt} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{6}} ^{\mathrm{10}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{10}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{6}}\mathrm{t}^{\mathrm{k}−\mathrm{6}} \right]_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \:\:+\mathrm{C}_{\mathrm{10}} ^{\mathrm{6}} \:\left[\mathrm{ln}\mid\mathrm{t}\mid\right]_{\mathrm{1}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{6}} ^{\mathrm{10}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\frac{\mathrm{C}_{\mathrm{10}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{6}}\left\{\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{k}−\mathrm{6}} −\mathrm{1}\right\}\:+\mathrm{C}_{\mathrm{10}} ^{\mathrm{6}} \:\mathrm{ln}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{11}} }\:\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{6}} ^{\mathrm{10}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{10}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{6}}\left\{\:\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{k}−\mathrm{6}} −\mathrm{1}\right\}+\frac{\mathrm{C}_{\mathrm{10}} ^{\mathrm{6}} }{\mathrm{5}^{\mathrm{11}} }\mathrm{ln}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\:. \\ $$

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