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Question Number 103742 by mathmax by abdo last updated on 17/Jul/20
calculate ∫_0 ^∞     (dx/((2x+1)^4 (x+3)^5 ))
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} } \\ $$
Answered by mathmax by abdo last updated on 17/Jul/20
A=∫_0 ^∞     (dx/((2x+1)^4 (x+3)^5 )) ⇒A =∫_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)/((2−t)^2 )) =(5/((2−t)^2 )) and x+3 =((3t−1)/(2−t)) +3 =((3t−1+6−3t)/(2−t)) =(5/(2−t))  ⇒A =∫_(1/3) ^2    (5/((2−t)^2  t^4 ((5/(2−t)))^9 )) dt =(1/5^8 ) ∫_(1/3) ^2  (((2−t)^9 )/((2−t)^2  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_(k=0) ^7  C_7 ^k (−2)^(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  [ln∣t∣]_(1/3) ^2 }  A=−(1/5^8 ){ Σ_(k=0 and k≠3) ^7  (−2)^(7−k )  (C_7 ^k /(k−3)){2^(k−3) −((1/3))^(k−3) }+(−2)^4  C_7 ^3 (ln2+ln3))}
$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} }\:\Rightarrow\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{9}} }\:\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}\:=\mathrm{t}\:\Rightarrow\mathrm{2x}+\mathrm{1}\:=\mathrm{tx}\:+\mathrm{3t}\:\Rightarrow\left(\mathrm{2}−\mathrm{t}\right)\mathrm{x}\:=\mathrm{3t}−\mathrm{1}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}\:\:=\frac{\mathrm{3}\left(\mathrm{2}−\mathrm{t}\right)+\mathrm{3t}−\mathrm{1}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }\:=\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }\:\mathrm{and}\:\mathrm{x}+\mathrm{3}\:=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}\:+\mathrm{3}\:=\frac{\mathrm{3t}−\mathrm{1}+\mathrm{6}−\mathrm{3t}}{\mathrm{2}−\mathrm{t}}\:=\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}} \\ $$$$\Rightarrow\mathrm{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\:\:\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{4}} \left(\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}}\right)^{\mathrm{9}} }\:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{9}} }{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{7}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}\:} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\mathrm{t}^{\mathrm{k}} \left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} \:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\mathrm{t}^{\mathrm{k}−\mathrm{4}\:} \mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\left\{\:\sum_{\mathrm{k}=\mathrm{0and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{3}}\mathrm{t}^{\mathrm{k}−\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\:+\left(−\mathrm{2}\right)^{\mathrm{4}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \:\left[\mathrm{ln}\mid\mathrm{t}\mid\right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \right\} \\ $$$$\left.\mathrm{A}=−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\left\{\:\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\left(−\mathrm{2}\right)^{\mathrm{7}−\mathrm{k}\:} \:\frac{\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{3}}\left\{\mathrm{2}^{\mathrm{k}−\mathrm{3}} −\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{k}−\mathrm{3}} \right\}+\left(−\mathrm{2}\right)^{\mathrm{4}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \left(\mathrm{ln2}+\mathrm{ln3}\right)\right)\right\} \\ $$

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