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Question Number 87993 by abdomathmax last updated on 07/Apr/20
calculate ∫_0 ^∞ (dx/((x+1)^2 (x+2)^2 (x+3)^2 ))
calculate∫0∞dx(x+1)2(x+2)2(x+3)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...
I=∫0∞dx(x+1)2(x+2)2(x+3)2⇒I=∫0∞dx(x+1x+2)2(x+2)4(x+3)2weusethechangementx+1x+2=t⇒x+1=tx+2t⇒(1−t)x=2t−1x=2t−11−t⇒dxdt=2(1−t)+(2t−1)(1−t)2=−3(t−1)2alsox+3=2t−11−t+3=2t−1+3−3t1−t=−t+21−t=t−2t−1⇒x+2=2t−11−t+2=2t−1+2−2t1−t=11−t⇒I=∫121−3dt(t−1)2t2(1t−1)4(t−2t−1)2=−3∫121(t−1)6(t−1)2t2(t−2)2dt=−3∫121(t−1)4t2(t−2)2dtwehave1t2(t−2)2=14(t−(t−2))2t2(t−2)2=14×t2−2t(t−2)+(t−2)2t2(t−2)2=14(t−2)2−12t(t−2)+14t2=14(t−2)2−14t−(t−2)t(t−2)+14t2=14(t−2)2−14(t−2)+14t+14t2⇒I=−34∫121(t−1)4(t−2)2dt+34∫121(t−1)4t−2dt−34∫121(t−1)4tdt−34∫121(t−1)4t2dt∫121(t−1)4(t−2)2dt=t−2=u∫−32−1(u−3)4u2du=∫−32−1∑k=04C4kuk(−3)4−ku2du=∑k=04(−3)kC4k∫−32−1uk−2du=∑k=0andk≠14(−3)kC4k[1k−1uk−1]−32−1−3C41[ln∣u∣]−32−1=∑k=0andk≠14(−3)kC4kk−1{(−1)k−1−(−32)k−1}−3C41(−ln(32))∫121(t−1)4t−2dt=t−2=u∫−32−1(u−3)4udu=∫−32−1∑k=04C4kuk(−3)4−kudu=∑k=04(−3)kC4k∫−32−1uk−1du=∑k=14(−3)kC4k[1kuk]−32−1+C40[ln∣u∣]−32−1=∑k=14(−3)kC4kk{(−1)k−(−32)k}−ln(32)weusethesamemethodtofindtheotherintegrals…

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