Paradox in indefinite integral Today,I met a problem about calculating such an indefinite integral$$ I=\int \frac{1}{(1+x^2)(1+x^{2019})}$$ My thought is as follows $$
\begin{aligned} I & = \int \frac { 1 } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = \int \frac { 1 } { \left( 1 + x ^ { - 2 } \right) \left( 1 + x ^ { - 2019} \right) } \left( - \frac { 1 } { x ^ { 2 } } \right) d x \\ & = \int \frac { - x ^ { 2019 } } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = I - \int \frac { 1 } { 1 + x ^ { 2 } } d x \end{aligned}
$$ No doubt,it is wrong.However, I don’t know how to illustrate this paradox?

Let $f_n$ denote an antiderivative of $\frac{1}{(1+x^2)(1+x^n)},\,n:=2019$ so $\int\frac{-x^ndx}{(1+x^2)(1+x^n)}=f_n^\prime(1/x)+C$. So you've actually shown $f_n(x)=f_n(1/x)-\arctan x+C_n$ for some $C_n$.
