Question Number 114056 by mathmax by abdo last updated on 16/Sep/20

calculate ∫_2 ^(+∞)     (dt/((2t+3)^4 (t−1)^5 ))

Answered by Olaf last updated on 17/Sep/20

R(x) =   ((1/(625))/((x−1)^5 ))  −((8/(3125))/((x−1)^4 ))  +((8/(3135))/((x−1)^3 ))  +(((32)/(15625))/((x−1)^2 ))  +(((112)/(78125))/(x−1))  +(((32)/(3125))/((2x+3)^4 ))  −(((32)/(3125))/((2x+3)^3 ))  −(((96)/(15625))/((2x+3)^2 ))  −(((225)/(78125))/(2x+3))  ∫_2 ^∞ R(x)dx = ...

Answered by Olaf last updated on 17/Sep/20

u = t−1  I = ∫_1 ^∞ (du/(u^5 (u+5)^4 ))  x = (1/u)  I = ∫_1 ^0 (x^5 /(((1/x)+5)^5 ))(−(dx/x^2 ))  I = ∫_0 ^1 (x^8 /((5x+1)^5 ))dx  λ = 5x+1  I = (1/5^8 )∫_1 ^6 (((λ−1)^8 )/λ^5 ).(dλ/5)  I = (1/5^9 )∫_1 ^6 ((Σ_(k=0) ^8 C_8 ^k λ^k (−1)^(8−k) )/λ^5 )dλ  ... may be this way is better...

Answered by 1549442205PVT last updated on 17/Sep/20

Put u=(1/(t−1))⇒du=((−1)/((t−1)^2 ))dt  dt=−(t−1)^2 du=((−1)/u^2 )du,t=(1/u)+1  ∫_2 ^∞     (dt/((2t+3)^4 (t−1)^5 ))=∫_1 ^0 ((−u^3 du)/(((2/u)+5)^4 ))  =−∫_1 ^0 (u^7 /((5u+2)^4 ))du=−(1/(625))∫_(7/5) ^0 (u^7 /((u+(2/5))^4 ))du  Put u+(2/5)=v⇒du=dv,u=v−(2/5)  I=−(1/(625))∫_(7/5) ^(2/5) (((v−(2/5))^7 )/v^4 )dv  =((−1)/(625))∫_(7/5) ^(2/5) {[v^7 −7v^6 .0.4+21v^5 .0.4^2   −35v^4 .0.4^3 +35v^3 .0.4^4 −21v^2 .(0.4)^5   +7v.(0.4)^6 −(0.4)^7 ]/v^4 }dv  =((−1)/(625))∫_(7/5) ^∞ [v^3 −((14)/5)v^2 +21v.((2/5))^2 −((35.8)/(125))  +((35.16)/(625v))−((21.32)/(3125v^2 ))+((7.64)/(15625v^3 ))−((128)/(5^7 v^4 ))]  =((−1)/(625)){(v^4 /4)−((14v^3 )/(15))+((42)/(25))v^2 −((280)/(125))v  +((560)/(625)).ln∣v∣+((672)/(3125v))−((224)/(15625v^2 ))+((128)/(3.5^7 v^3 ))]_(7/5) ^(2/5)   =−(1/(625))[(−1.0449965−(−0.99590280)]=  (−0.0490937)/625=0.00007855

Answered by MJS_new last updated on 17/Sep/20

∫(dt/((2t+3)^4 (t−1)^5 ))=       [Ostrogradski′s Method]  =((5376t^6 +1344t^5 −25760t^4 +3080t^3 +40740t^2 −17906t−16249)/(187500(t−1)^4 (2t+3)^3 ))+  +((112)/(15625))∫(dt/((t−1)(2t+3)))  this integral=((112)/(78125))ln ∣((t−1)/(2t+3))∣  ⇒ answer is ((112)/(78125))(ln 7 −ln 2)−((36817)/(21437500))

Commented by1549442205PVT last updated on 17/Sep/20

Great !Sir′s result coincde to my result

Commented byMJS_new last updated on 17/Sep/20

yes.  I like Ostrogradski′s Method better than  decomposing...