Question

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

Answer 1

Let $S_n = \sum\limits_{k=0}^n x^k$. Then

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

$xS_n = x + x^2 + x^3 + \cdots + x^{n+1}$

$\implies (1 - x) S_n = 1 - x^{n+1}$

$\implies S_n = \dfrac{1 - x^{n+1}}{1 - x}$

$\implies \displaystyle \sum_{k=0}^{2019} x^k = S_{2019} = \frac{1 - x^{2020}}{1 - x}$

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

$\displaystyle S_n' = \sum_{k=1}^n k x^k$

$\displaystyle S_n' = x \sum_{k=1}^n k x^{k-1}$

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

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

$\displaystyle S_n' = x (S_{n-1}' + S_{n-1})$

$\displaystyle S_n' = x (S_n' - nx^n) + xS_{n-1}$

$\implies S_n' = \dfrac{x - (n+1)x^{n+1} + nx^{n+2}}{(1-x)^2}$

$\implies \displaystyle \sum_{k=1}^{2018} kx^k = S_{2018}' = \dfrac{x - 2019x^{2019} + 2018x^{2020}}{(1-x)^2}$

So, the integrand reduces considerably to

$\dfrac1{x-1} + \dfrac{S_{2018}' + S_{2018}}{S_{2019}} = \dfrac{2020 x^{2019}}{x^{2020}-1}$

and its integral is trivial; by substitution,

$\displaystyle \int \frac{2020x^{2019}}{x^{2020}-1} \, dx = \int \frac{d(x^{2020}-1)}{x^{2020}-1} = \boxed{\ln\left|x^{2020}-1\right| + C}$