Question

`\rm\begin{gathered}\displaystyle\int \rm \left[ \rm (1)/(x - 1) + ( \displaystyle\rm \sum_(k = 0)^(2018) (k + 1)x^k)/(\displaystyle \rm\sum_(k = 0)^(2019)x^k) \right]dx \\\end{gathered}`​

Answer 1

Let
`S_n = \sum\limits_(k=0)^n x^k`. Then

`S_n = 1 + x + x^2 + \cdots + x^n`

`xS_n = x + x^2 + x^3 + \cdots + x^(n+1)`

`\implies (1 - x) S_n = 1 - x^(n+1)`

`\implies S_n = (1 - x^(n+1))/(1 - x)`

`\implies \displaystyle \sum_(k=0)^(2019) x^k = S_(2019) = (1 - x^(2020))/(1 - x)`

Let
`S_n' = \sum\limits_(k=1)^n k x^k`. (Starting the sum at k = 0 doesn't change its value.) Then

`\displaystyle S_n' = \sum_(k=1)^n k x^k`

`\displaystyle S_n' = x \sum_(k=1)^n k x^(k-1)`

`\displaystyle S_n' = x \sum_(k=0)^(n-1) (k+1) x^k`

`\displaystyle S_n' = x \left(\sum_(k=0)^(n-1) kx^k + \sum_(k=0)^(n-1) x^k\right)`

`\displaystyle S_n' = x (S_(n-1)' + S_(n-1))`

`\displaystyle S_n' = x (S_n' - nx^n) + xS_(n-1)`

`\implies S_n' = (x - (n+1)x^(n+1) + nx^(n+2))/((1-x)^2)`

`\implies \displaystyle \sum_(k=1)^(2018) kx^k = S_(2018)' = (x - 2019x^(2019) + 2018x^(2020))/((1-x)^2)`

So, the integrand reduces considerably to

`\frac1{x-1} + (S_(2018)' + S_(2018))/(S_(2019)) = (2020 x^(2019))/(x^(2020)-1)`

and its integral is trivial; by substitution,

`\displaystyle \int (2020x^(2019))/(x^(2020)-1) \, dx = \int (d(x^(2020)-1))/(x^(2020)-1) = \boxed\ln\left`