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Question Number 124531 by Bird last updated on 03/Dec/20

calculate ∫_0 ^∞  (dx/((2x+1)^4 (x+3)^5 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}+\mathrm{3}\right)^{\mathrm{5}} } \\ $$

Answered by mathmax by abdo last updated on 04/Dec/20

let I =∫_0 ^∞  (dx/((2x+1)^4 (x+3)^5 )) ⇒I =∫_0 ^∞  (dx/((((2x+1)/(x+3)))^4 (x+3)^9 ))  we do the changement ((2x+1)/(x+3))=t ⇒2x+1=tx+3t ⇒(2−t)x=3t−1  ⇒x=((3t−1)/(2−t)) ⇒(dx/dt)=((3(2−t)−(3t−1)(−1))/((2−t)^2 ))=((6−3t+3t−1)/((2−t)^2 ))=(5/((2−t)^2 ))  and x+3=((3t−1)/(2−t))+3 =((3t−1+6−3t)/(2−t))=(5/(2−t)) ⇒  I =∫_(1/3) ^2  (5/((2−t)^2 t^4 ((5/(2−t)))^9 ))dt =(1/5^8 )∫_(1/3) ^2  (((2−t)^9 )/(t^4 ((2−t)^2 ))dt  =(1/5^8 )∫_(1/3) ^2   (((2−t)^7 )/t^4 )dt =−(1/5^8 )∫_(1/3) ^2  (((t−2)^7 )/t^4 )dt  =−(1/5^8 )∫_(1/3) ^2   ((Σ_(k=0) ^7  C_7 ^k  t^k (−2)^(7−k) )/t^4 )dt  =−(1/5^8 )Σ_(k0) ^7  (−2)^(7−k)  C_7 ^k   ∫_(1/3) ^2  t^(k−4)  dt  =−(1/5^8 ) {Σ_(k=0and k≠3) ^7  (−2)^(7−k)  C_7 ^k   [(1/(k−3))t^(k−3) ]_(1/3) ^2   +(−2)^4 C_7 ^3 [ln2−ln((1/3)))  I=−(1/5^8 )Σ_(k=0 and ≠3) ^7   (−2)^(7−k)  (C_7 ^k /(k−3)){2^(k−3) −(1/3^(k−3) )}  +16 C_7 ^3 ln(6)

$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} }\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{9}} } \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}=\mathrm{t}\:\Rightarrow\mathrm{2x}+\mathrm{1}=\mathrm{tx}+\mathrm{3t}\:\Rightarrow\left(\mathrm{2}−\mathrm{t}\right)\mathrm{x}=\mathrm{3t}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{3}\left(\mathrm{2}−\mathrm{t}\right)−\left(\mathrm{3t}−\mathrm{1}\right)\left(−\mathrm{1}\right)}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }=\frac{\mathrm{6}−\mathrm{3t}+\mathrm{3t}−\mathrm{1}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }=\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{x}+\mathrm{3}=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}+\mathrm{3}\:=\frac{\mathrm{3t}−\mathrm{1}+\mathrm{6}−\mathrm{3t}}{\mathrm{2}−\mathrm{t}}=\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{4}} \left(\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}}\right)^{\mathrm{9}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{9}} }{\mathrm{t}^{\mathrm{4}} \left(\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} \right.}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\:\frac{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{7}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{7}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\mathrm{t}^{\mathrm{k}} \left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\sum_{\mathrm{k0}} ^{\mathrm{7}} \:\left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\mathrm{t}^{\mathrm{k}−\mathrm{4}} \:\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\:\left\{\sum_{\mathrm{k}=\mathrm{0and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{3}}\mathrm{t}^{\mathrm{k}−\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\:+\left(−\mathrm{2}\right)^{\mathrm{4}} \mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \left[\mathrm{ln2}−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)\right. \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\neq\mathrm{3}} ^{\mathrm{7}} \:\:\left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} \:\frac{\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{3}}\left\{\mathrm{2}^{\mathrm{k}−\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{k}−\mathrm{3}} }\right\} \\ $$$$+\mathrm{16}\:\mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \mathrm{ln}\left(\mathrm{6}\right) \\ $$

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