Question

If $\int \frac{x^{2018}}{1+x+\frac{x^2}{2!}+\cdots +\frac{x^{2018}}{2018!}}dx=m!x-m!\ln |P\left(x\right)|+C$ for arbitrary constant of integration $C,$ then

• A. $P\left(x\right)=1+x+\frac{x^2}{2!}+\cdots +\frac{x^{2018}}{2018!}$
• B. $P\left(x\right)=x+\frac{x^2}{2!}+\cdots +\frac{x^{m+1}}{(m+1)!}$
• C. $m=2019$
• D. $m=2018$

Answer 1

Answer: (A), (D)

$m=2018$

Let $f\left(x\right)=1+x+\frac{x^2}{2!}+\cdots +\frac{x^{2018}}{2018!}$
$f^{\prime }\left(x\right)=1+\frac{x}{1!}+\cdots +\frac{x^{2017}}{2017!}$

$\int \frac{x^{2018}}{1+x+\frac{x^2}{2!}+\cdots +\frac{x^{2018}}{2018!}}dx$
$=\int \left(2018\right)!\frac{\left(f\right(x)-f^{\prime }(x\left)\right)}{f\left(x\right)}dx$
$=\left(2018\right)!\int (1-\frac{f^{\prime }\left(x\right)}{f\left(x\right)})dx$
$=2018!(x-\ln |f(x\left)|\right)+C$
Comparing with $m!x-m!\ln |P\left(x\right)|+C,$
$m=2018$ and $P\left(x\right)=1+x+\frac{x^2}{2!}+\cdots +\frac{x^{2018}}{2018!}$