Question Number 113929 by gloriousman last updated on 16/Sep/20

∫(x+1)^2 (1−x)^5 dx

Commented byShakaLaka last updated on 16/Sep/20

we can also solve by simplification i.e., ∫(x+1)^2 (1−x)^5 dx =∫(x^2 +2x+1)(1−x)^3 (1−x)^2 dx =∫(x^2 +2x+1)(1−3x+3x^2 −x^3 ) (x^2 −2x+1) dx =∫(x^2 +2x+1−3x^3 −6x^2 −3x +3x^4 +6x^3 +3x^2 −x^5 −2x^4 −x^3 ) (x^2 −2x+1) dx =∫(−x^5 +x^4 +2x^3 −2x^2 −x+1) (x^2 −2x+1) dx =∫(−x^7 +x^6 +2x^5 −2x^4 −x^3 +x^2 +2x^6 −2x^5 −4x^4 +4x^3 +2x^2 −2x −x^5 +x^4 +2x^3 −2x^2 −x+1) dx =∫(−x^7 +3x^6 −x^5 −5x^4 +5x^3 +x^2 −3x+1) dx =−(1/8)x^8 +(3/7)x^7 −(1/6)x^6 −x^5 +(5/4)x^4 +(1/3)x^3 −(3/2)x^2 +x+C

Commented byMJS_new last updated on 16/Sep/20

yes we can. but try ∫(x+3)^2 (x−7)^(100) dx

Commented byShakaLaka last updated on 16/Sep/20

yeah, you are absolutely right

Commented bymathmax by abdo last updated on 16/Sep/20

we get the answer by variable changement A =∫ (x+3)^2 (x−7)^(100) dx =_(x−7 =t) ∫ t^(100) (t+10)^2 dt =∫ t^(100) (t^2 +20t +100)dt=∫ (t^(102) +20t^(101) +100 t^(100) )dt =(1/(103))t^(103) +((20)/(102)) t^(102) +((100)/(101)) t^(101) +C

Answered by MJS_new last updated on 16/Sep/20

∫(x+1)^2 (1−x)^5 dx= [t=1−x → dx=dt] =∫t^5 (t−2)^2 dt=∫t^7 −4t^6 +4t^5 dt= =(1/8)t^8 −(4/7)t^7 +(2/3)t^6 =((t^6 (21t^2 −96t+112))/(168))= =(((1−x)^6 (21x^2 +54x+37))/(168))+C

Answered by 1549442205PVT last updated on 16/Sep/20

(x+1)^2 (1−x)^5 =[(1−x)(1+x)]^2 (1−x)^3 =(1−x^2 )^2 (1−x)^3 =(x^4 −2x^2 +1)(1−3x+3x^2 −x^3 ) =x^4 −2x^2 +1−3x^5 +6x^3 −3x+3x^6 −6x^4 +3x^2 −x^7 +2x^5 −x^3 =−x^7 +3x^6 −x^5 −5x^4 +5x^3 +x^2 −3x+1 Hence,∫(x+1)^2 (1−x)^5 dx =∫(−x^7 +3x^6 −x^5 −5x^4 +5x^3 +x^2 −3x+1)dx =((−x^8 )/8)+((3x^7 )/7)−(x^6 /6)−x^5 +((5x^4 )/4)+(x^3 /3)−((3x^2 )/2)+x+C

Commented byMJS_new last updated on 16/Sep/20

yes. but in this special case and similar ones the path I went is shorter. try ∫(x+3)^2 (x−7)^(100) dx

Commented by1549442205PVT last updated on 16/Sep/20

Thank Sir.