Question Number 87993 by abdomathmax last updated on 07/Apr/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 07/Apr/20
![I =∫_0 ^∞ (dx/((x+1)^2 (x+2)^2 (x+3)^2 )) ⇒I =∫_0 ^∞ (dx/((((x+1)/(x+2)))^2 (x+2)^4 (x+3)^2 )) we use the changement ((x+1)/(x+2)) =t ⇒x+1 =tx +2t ⇒(1−t)x=2t−1 x =((2t−1)/(1−t)) ⇒(dx/dt) =((2(1−t)+(2t−1))/((1−t)^2 )) =((−3)/((t−1)^2 )) also x+3 =((2t−1)/(1−t)) +3 =((2t−1+3−3t)/(1−t)) =((−t+2)/(1−t)) =((t−2)/(t−1)) ⇒ x+2 =((2t−1)/(1−t))+2 =((2t−1+2−2t)/(1−t)) =(1/(1−t)) ⇒ I =∫_(1/2) ^1 ((−3dt)/((t−1)^2 t^2 ((1/(t−1)))^4 (((t−2)/(t−1)))^2 )) =−3∫_(1/2) ^1 (((t−1)^6 )/((t−1)^2 t^2 (t−2)^2 ))dt =−3 ∫_(1/2) ^1 (((t−1)^4 )/(t^2 (t−2)^2 ))dt we have (1/(t^2 (t−2)^2 )) =(1/4)(((t−(t−2))^2 )/(t^2 (t−2)^2 )) =(1/4)×((t^2 −2t(t−2)+(t−2)^2 )/(t^2 (t−2)^2 )) =(1/(4(t−2)^2 ))−(1/(2t(t−2))) +(1/(4t^2 )) =(1/(4(t−2)^2 ))−(1/4)((t−(t−2))/(t(t−2)))+(1/(4t^2 )) =(1/(4(t−2)^2 ))−(1/(4(t−2))) +(1/(4t))+(1/(4t^2 )) ⇒ I =−(3/4)∫_(1/2) ^1 (((t−1)^4 )/((t−2)^2 ))dt +(3/4)∫_(1/2) ^1 (((t−1)^4 )/(t−2))dt−(3/4)∫_(1/2) ^1 (((t−1)^4 )/t)dt−(3/4)∫_(1/2) ^1 (((t−1)^4 )/t^2 )dt ∫_(1/2) ^1 (((t−1)^4 )/((t−2)^2 ))dt =_(t−2 =u) ∫_(−(3/2)) ^(−1) (((u−3)^4 )/u^2 )du =∫_(−(3/2)) ^(−1) ((Σ_(k=0) ^4 C_4 ^k u^k (−3)^(4−k) )/u^2 )du =Σ_(k=0) ^4 (−3)^k C_4 ^k ∫_(−(3/2)) ^(−1) u^(k−2) du =Σ_(k=0and k≠1) ^4 (−3)^k C_4 ^k [(1/(k−1))u^(k−1) ]_(−(3/2)) ^(−1) −3 C_4 ^1 [ln∣u∣]_(−(3/2)) ^(−1) =Σ_(k=0 and k≠1) ^4 (−3)^k (C_4 ^k /(k−1)){ (−1)^(k−1) −(−(3/2))^(k−1) } −3C_4 ^1 (−ln((3/2))) ∫_(1/2) ^1 (((t−1)^4 )/(t−2))dt =_(t−2 =u) ∫_(−(3/2)) ^(−1) (((u−3)^4 )/u)du =∫_(−(3/2)) ^(−1) ((Σ_(k=0) ^4 C_4 ^k u^k (−3)^(4−k) )/u)du =Σ_(k=0) ^4 (−3)^k C_4 ^k ∫_(−(3/2)) ^(−1) u^(k−1) du =Σ_(k=1) ^4 (−3)^k C_4 ^k [(1/k) u^k ]_(−(3/2)) ^(−1) +C_4 ^0 [ln∣u∣]_(−(3/2)) ^(−1) =Σ_(k=1) ^4 (−3)^k (C_4 ^k /k){ (−1)^k −(−(3/2))^k }−ln((3/2)) we use the same method to find the other integrals...](https://www.tinkutara.com/question/Q88031.png)
$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({x}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{4}} \left({x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${we}\:{use}\:{the}\:{changement}\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\:={t}\:\Rightarrow{x}+\mathrm{1}\:={tx}\:+\mathrm{2}{t}\:\Rightarrow\left(\mathrm{1}−{t}\right){x}=\mathrm{2}{t}−\mathrm{1} \\ $$$${x}\:=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{2}\left(\mathrm{1}−{t}\right)+\left(\mathrm{2}{t}−\mathrm{1}\right)}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{3}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:\:{also} \\ $$$${x}+\mathrm{3}\:=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{1}−{t}}\:+\mathrm{3}\:=\frac{\mathrm{2}{t}−\mathrm{1}+\mathrm{3}−\mathrm{3}{t}}{\mathrm{1}−{t}}\:=\frac{−{t}+\mathrm{2}}{\mathrm{1}−{t}}\:=\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\:\Rightarrow \\ $$$${x}+\mathrm{2}\:=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{1}−{t}}+\mathrm{2}\:=\frac{\mathrm{2}{t}−\mathrm{1}+\mathrm{2}−\mathrm{2}{t}}{\mathrm{1}−{t}}\:=\frac{\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow \\ $$$${I}\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{−\mathrm{3}{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{t}−\mathrm{1}}\right)^{\mathrm{4}} \left(\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\right)^{\mathrm{2}} }\:=−\mathrm{3}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{6}} }{\left({t}−\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)^{\mathrm{2}} }{dt} \\ $$$$=−\mathrm{3}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)^{\mathrm{2}} }{dt}\:\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\frac{\left({t}−\left({t}−\mathrm{2}\right)\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}×\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}\left({t}−\mathrm{2}\right)+\left({t}−\mathrm{2}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{t}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\frac{{t}−\left({t}−\mathrm{2}\right)}{{t}\left({t}−\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}{t}}+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{3}}{\mathrm{4}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{\left({t}−\mathrm{2}\right)^{\mathrm{2}} }{dt}\:+\frac{\mathrm{3}}{\mathrm{4}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}−\mathrm{2}}{dt}−\frac{\mathrm{3}}{\mathrm{4}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}}{dt}−\frac{\mathrm{3}}{\mathrm{4}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}^{\mathrm{2}} }{dt} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{\left({t}−\mathrm{2}\right)^{\mathrm{2}} }{dt}\:=_{{t}−\mathrm{2}\:={u}} \:\:\:\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\frac{\left({u}−\mathrm{3}\right)^{\mathrm{4}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{C}_{\mathrm{4}} ^{{k}} \:{u}^{{k}} \left(−\mathrm{3}\right)^{\mathrm{4}−{k}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \:\:\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} {u}^{{k}−\mathrm{2}} \:{du} \\ $$$$=\sum_{{k}=\mathrm{0}{and}\:{k}\neq\mathrm{1}} ^{\mathrm{4}} \:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \:\:\left[\frac{\mathrm{1}}{{k}−\mathrm{1}}{u}^{{k}−\mathrm{1}} \right]_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\:\:−\mathrm{3}\:{C}_{\mathrm{4}} ^{\mathrm{1}} \:\left[{ln}\mid{u}\mid\right]_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{1}} ^{\mathrm{4}} \:\left(−\mathrm{3}\right)^{{k}} \:\frac{{C}_{\mathrm{4}} ^{{k}} }{{k}−\mathrm{1}}\left\{\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} −\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)^{{k}−\mathrm{1}} \right\} \\ $$$$−\mathrm{3}{C}_{\mathrm{4}} ^{\mathrm{1}} \left(−{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}−\mathrm{2}}{dt}\:=_{{t}−\mathrm{2}\:={u}} \:\:\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\:\frac{\left({u}−\mathrm{3}\right)^{\mathrm{4}} }{{u}}{du} \\ $$$$=\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} {C}_{\mathrm{4}} ^{{k}} \:{u}^{{k}} \left(−\mathrm{3}\right)^{\mathrm{4}−{k}} }{{u}}{du} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \:\:\int_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:{u}^{{k}−\mathrm{1}} \:{du} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{4}} \left(−\mathrm{3}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \left[\frac{\mathrm{1}}{{k}}\:{u}^{{k}} \right]_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \:\:+{C}_{\mathrm{4}} ^{\mathrm{0}} \:\left[{ln}\mid{u}\mid\right]_{−\frac{\mathrm{3}}{\mathrm{2}}} ^{−\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{4}} \:\left(−\mathrm{3}\right)^{{k}} \:\frac{{C}_{\mathrm{4}} ^{{k}} }{{k}}\left\{\:\left(−\mathrm{1}\right)^{{k}} −\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)^{{k}} \right\}−{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${we}\:{use}\:{the}\:{same}\:{method}\:{to}\:{find}\:{the}\:{other}\:{integrals}… \\ $$