calculate-2019-2021-dx-x-1-2019-x-1-2021-

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 = \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}}

\dots

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 = \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})}