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Evaluate-0-x-1-2020-x-2-2022-dx-

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Question Number 166756 by HongKing last updated on 27/Feb/22
Evaluate: ∫_0 ^( ∞) (((x + 1)^(2020) )/((x + 2)^(2022) )) dx = ?
Evaluate:∫0∞(x+1)2020(x+2)2022dx=?
Answered by Mathspace last updated on 27/Feb/22
x+2=t ⇒I=∫_2 ^∞ (((t−1)^(2020) )/t^(2022) )dt =∫_2 ^∞ ((Σ_(k=0) ^(2020) C_(2020) ^k t^k (−1)^(2020−k) )/t^(2022) )dt =Σ_(k=0) ^(2020) (−1)^k C_(2020) ^k t^(k−2022) dt =Σ_(k=0) ^(2020) (−1)^k C_(2020) ^k [(1/(k−2021))t^(k−2022) ]_2 ^∞ =−Σ_(k=0) ^(2020) (((−1)^k C_(2020) ^k )/(k−2021)) 2^(k−2022)
x+2=t⇒I=∫2∞(t−1)2020t2022dt=∫2∞∑k=02020C2020ktk(−1)2020−kt2022dt=∑k=02020(−1)kC2020ktk−2022dt=∑k=02020(−1)kC2020k[1k−2021tk−2022]2∞=−∑k=02020(−1)kC2020kk−20212k−2022
Answered by mathsmine last updated on 27/Feb/22
=∫_0 ^∞ (((x+1)/(x+2)))^(2020) .((1/(x+2)))^2 dx u=((x+1)/(x+2))=1−(1/((x+2)))⇒du=(dx/((x+2)^2 )) =∫_(1/2) ^1 u^(2020) du=(1/(2021))(1−(1/2^(2021) ))
=∫0∞(x+1x+2)2020.(1x+2)2dxu=x+1x+2=1−1(x+2)⇒du=dx(x+2)2=∫121u2020du=12021(1−122021)

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