calculate ∫_(2019) ^(2021) (dx/((x−1)^(2019) (x+1)^(2021) ))


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Question Number 126873 by mathmax by abdo last updated on 25/Dec/20

$calculate \int_{2019}^{2021} \frac{dx}{{(x - 1)}^{2019} {(x + 1)}^{2021}}$
	Answered by Ar Brandon last updated on 25/Dec/20
	$I=∫_(2019) ^(2021) (dx/((x−1)^(2019) (x+1)^(2021) )) =∫_(2019) ^(2021) (((x−1)^2 )/((x−1)^(2021) (x+1)^(2021) ))dx =∫_(2019) ^(2021) ((x^2 −2x+1)/((x^2 −1)^(2021) ))dx x^2 −2x+1=λ(x^2 −1)+μ{(d/dx)(x^2 −1)}+γ =λ(x^2 −1)+μ(2x)+γ λ=1, μ=−1, −λ+γ=1, γ=2 I=∫_(2019) ^(2021) {((x^2 −1)/((x^2 −1)^(2021) ))−((2x)/((x^2 −1)^(2021) ))+(2/((x^2 −1)^(2021) ))}dx =[(1/((x^2 −1)^(2020) ))]_(2019) ^(2021) +∫_(2019) ^(2021) {((x^2 −1)/((x^2 −1)^(2021) ))+(2/((x^2 −1)^(2021) ))}dx f(a)=∫(dx/(x^2 −a^2 ))=−(1/a)Arctanh((x/a))+C f ′(a)=∫((2a)/((x^2 −a^2 )^2 )) ...$
	$I = \int_{2019}^{2021} \frac{dx}{{(x - 1)}^{2019} {(x + 1)}^{2021}}$ $= \int_{2019}^{2021} \frac{{(x - 1)}^{2}}{{(x - 1)}^{2021} {(x + 1)}^{2021}} dx$ $= \int_{2019}^{2021} \frac{x^{2} - 2 x + 1}{{(x^{2} - 1)}^{2021}} dx$ $x^{2} - 2 x + 1 = λ (x^{2} - 1) + μ {\frac{d}{dx} (x^{2} - 1)} + γ$ $= λ (x^{2} - 1) + μ (2 x) + γ$ $λ = 1, μ = - 1, - λ + γ = 1, γ = 2$ $I = \int_{2019}^{2021} {\frac{x^{2} - 1}{{(x^{2} - 1)}^{2021}} - \frac{2 x}{{(x^{2} - 1)}^{2021}} + \frac{2}{{(x^{2} - 1)}^{2021}}} dx$ $= {[\frac{1}{{(x^{2} - 1)}^{2020}}]}_{2019}^{2021} + \int_{2019}^{2021} {\frac{x^{2} - 1}{{(x^{2} - 1)}^{2021}} + \frac{2}{{(x^{2} - 1)}^{2021}}} dx$ $f (a) = \int \frac{dx}{x^{2} - a^{2}} = - \frac{1}{a} Arctanh (\frac{x}{a}) + C$ $f^{'} (a) = \int \frac{2 a}{{(x^{2} - a^{2})}^{2}}$ $. . .$
	Answered by mindispower last updated on 25/Dec/20

	$= \int_{2019}^{2021} \frac{d x}{{(x + 1)}^{2} {(\frac{x - 1}{x + 1})}^{2019}}$ $t = \frac{x - 1}{x + 1} \Rightarrow d t = \frac{2}{{(x + 1)}^{2}} d x$ $\Leftrightarrow \frac{1}{2} \int_{\frac{2018}{2020}}^{\frac{2020}{2022}} \frac{d t}{t^{2019}} = \frac{1}{2 (- 2018)} [{(\frac{2020}{2022})}^{- 2018} - {(\frac{2018}{2020})}^{- 2018}]$
	Commented by mathmax by abdo last updated on 27/Dec/20

	$not correct sir mind!$
	Answered by mathmax by abdo last updated on 27/Dec/20
	$I =∫_(2019) ^(2021) (dx/((x−1)^(2019) (x+1)^(2021) )) ⇒I =∫_(2019) ^(2021) (dx/((((x−1)/(x+1)))^(2019) (x+1)^(2021+2019) )) we do the changement ((x−1)/(x+1))=t ⇒x−1=tx+t ⇒(1−t)x=t+1 ⇒ x=((1+t)/(1−t)) ⇒(dx/dt)=((1−t−(1+t)(−1))/((1−t)^2 ))=(2/((1−t)^2 )) and x+1=((1+t)/(1−t))+1=((1+t+1−t)/(1−t))=(2/(1−t)) I =∫_(−((1010)/(1009))) ^(−((1011)/(1010))) (2/((1−t)^2 ×t^(2019) ((2/(1−t)))^(2021+2019) ))dt =(2/2^(4040) ) ∫_(−((1010)/(1009))) ^(−((1011)/(1010))) (((1−t)^(4040) )/((1−t)^2 t^(2019) ))dt =(1/2^(4039) )∫_(−((1010)/(1009))) ^(−((1011)/(1010))) (((t−1)^(4038) )/t^(2019) )dt =(1/2^(4039) )∫_(−((1010)/(1009))) ^(−((1011)/(1010))) ((Σ_(k=0) ^(4038) C_(4038) ^k t^k (−1)^(4038−k) )/t^(2019) )dt =(1/2^(4039) )Σ_(k=0) ^(4038) (−1)^k C_(4038) ^k ∫_(−((1010)/(1009))) ^(−((1011)/(1010))) t^(k−2019) dt =(1/2^(4039) ) Σ_(k=0 and k≠2018) ^(4038) (((−1)^k C_(4038) ^k )/(k−2018))[ t^(k−2018) ]_(−((1010)/(1009))) ^(−((1011)/(1010))) +(1/2^(4039) )C_(4038) ^(2018) {ln(((1011)/(1010)))−ln(((1010)/(1009)))} ⇒ I =(1/2^(4039) ) Σ_(k=0 andk≠2018) ^(4038) (((−1)^k C_(4038) ^k )/(k−2018)){(−((1011)/(1010)))^(k−2018) −(−((1010)/(1009)))^(k−2018) } +(1/2^(4039) ) C_(4038) ^(2018) {ln(((1011)/(1010)))−ln(((1010)/(1009)))}$
	$I = \int_{2019}^{2021} \frac{dx}{{(x - 1)}^{2019} {(x + 1)}^{2021}} \Rightarrow I = \int_{2019}^{2021} \frac{dx}{{(\frac{x - 1}{x + 1})}^{2019} {(x + 1)}^{2021 + 2019}}$ $we do the changement \frac{x - 1}{x + 1} = t \Rightarrow x - 1 = tx + t \Rightarrow (1 - t) x = t + 1 \Rightarrow$ $x = \frac{1 + t}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{1 - t - (1 + t) (- 1)}{{(1 - t)}^{2}} = \frac{2}{{(1 - t)}^{2}} and x + 1 = \frac{1 + t}{1 - t} + 1 = \frac{1 + t + 1 - t}{1 - t} = \frac{2}{1 - t}$ $I = \int_{- \frac{1010}{1009}}^{- \frac{1011}{1010}} \frac{2}{{(1 - t)}^{2} \times t^{2019} {(\frac{2}{1 - t})}^{2021 + 2019}} dt$ $= \frac{2}{2^{4040}} \int_{- \frac{1010}{1009}}^{- \frac{1011}{1010}} \frac{{(1 - t)}^{4040}}{{(1 - t)}^{2} t^{2019}} dt$ $= \frac{1}{2^{4039}} \int_{- \frac{1010}{1009}}^{- \frac{1011}{1010}} \frac{{(t - 1)}^{4038}}{t^{2019}} dt = \frac{1}{2^{4039}} \int_{- \frac{1010}{1009}}^{- \frac{1011}{1010}} \frac{\sum_{k = 0}^{4038} C_{4038}^{k} t^{k} {(- 1)}^{4038 - k}}{t^{2019}} dt$ $= \frac{1}{2^{4039}} \sum_{k = 0}^{4038} {(- 1)}^{k} C_{4038}^{k} \int_{- \frac{1010}{1009}}^{- \frac{1011}{1010}} t^{k - 2019} dt$ $= \frac{1}{2^{4039}} \sum_{k = 0 and k \neq 2018}^{4038} \frac{{(- 1)}^{k} C_{4038}^{k}}{k - 2018} {[t^{k - 2018}]}_{- \frac{1010}{1009}}^{- \frac{1011}{1010}}$ $+ \frac{1}{2^{4039}} C_{4038}^{2018} {\ln (\frac{1011}{1010}) - \ln (\frac{1010}{1009})} \Rightarrow$ $I = \frac{1}{2^{4039}} \sum_{k = 0 andk \neq 2018}^{4038} \frac{{(- 1)}^{k} C_{4038}^{k}}{k - 2018} {{(- \frac{1011}{1010})}^{k - 2018} - {(- \frac{1010}{1009})}^{k - 2018}}$ $+ \frac{1}{2^{4039}} C_{4038}^{2018} {\ln (\frac{1011}{1010}) - \ln (\frac{1010}{1009})}$
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