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1-x-1-k-0-2018-k-1-x-k-k-0-2019-x-k-dx-




Question Number 87723 by Ar Brandon last updated on 05/Apr/20
∫((1/(x−1))+((Σ_(k=0) ^(2018) (k+1)x^k )/(Σ_(k=0) ^(2019) x^k )))dx
$$\int\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\underset{{k}=\mathrm{0}} {\overset{\mathrm{2018}} {\sum}}\left({k}+\mathrm{1}\right){x}^{{k}} }{\underset{{k}=\mathrm{0}} {\overset{\mathrm{2019}} {\sum}}{x}^{{k}} }\right){dx} \\ $$
Answered by mahdi last updated on 05/Apr/20
 A=Σ_0 ^(2018) (i+1)x^i =1x^0 +2x^1 +...+2019^(2018) =  (x^1 +x^2 +...+x^(2019) )′=(((x^(2020) −x)/(x−1)))′  B=Σ_0 ^(2019) x^k =x^0 +x^1 +...+x^(2019) =((x^(2020) −1)/(x−1))  (A/B)=((((2020x^(2019) −1)(x−1)−(x^(2020) −x)(1))/((x−1)^2 ))/((x^(2020) −1)/(x−1)))  (A/B)=((2019x^(2020) −2020x^(2019) +1)/(x−1))  ∫((1/(x−1))+(A/B))dx=∫((2020x^(2019) )/(x^(2020) −1))dx  =ln(x^(2020) −1)
$$\:\mathrm{A}=\underset{\mathrm{0}} {\overset{\mathrm{2018}} {\sum}}\left(\mathrm{i}+\mathrm{1}\right)\mathrm{x}^{\mathrm{i}} =\mathrm{1x}^{\mathrm{0}} +\mathrm{2x}^{\mathrm{1}} +…+\mathrm{2019}^{\mathrm{2018}} = \\ $$$$\left(\mathrm{x}^{\mathrm{1}} +\mathrm{x}^{\mathrm{2}} +…+\mathrm{x}^{\mathrm{2019}} \right)'=\left(\frac{\mathrm{x}^{\mathrm{2020}} −\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)' \\ $$$$\mathrm{B}=\underset{\mathrm{0}} {\overset{\mathrm{2019}} {\sum}}\mathrm{x}^{\mathrm{k}} =\mathrm{x}^{\mathrm{0}} +\mathrm{x}^{\mathrm{1}} +…+\mathrm{x}^{\mathrm{2019}} =\frac{\mathrm{x}^{\mathrm{2020}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\frac{\left(\mathrm{2020x}^{\mathrm{2019}} −\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)−\left(\mathrm{x}^{\mathrm{2020}} −\mathrm{x}\right)\left(\mathrm{1}\right)}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }}{\frac{\mathrm{x}^{\mathrm{2020}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}}} \\ $$$$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\mathrm{2019x}^{\mathrm{2020}} −\mathrm{2020x}^{\mathrm{2019}} +\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$\int\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{A}}{\mathrm{B}}\right)\mathrm{dx}=\int\frac{\mathrm{2020x}^{\mathrm{2019}} }{\mathrm{x}^{\mathrm{2020}} −\mathrm{1}}\mathrm{dx} \\ $$$$=\mathrm{ln}\left(\mathrm{x}^{\mathrm{2020}} −\mathrm{1}\right) \\ $$
Commented by Ar Brandon last updated on 06/Apr/20
nice
$${nice} \\ $$

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