Two strange integrals

Question

I was asked to solve the two integrals

$$ \int\frac{1}{x^2e^{1/x}-x}dx $$ $$ \int\frac{x^2+1}{x^3-x^2+x+1}dx $$ I think the first one is not soluble with the known methods of calculus and the second one is involved with complex roots of cubic equation.

Any improvements appreciated.

Thanks.

Answer 1

1. For the integral

$$ \int\frac{x^2+1}{x^3-x^2+x+1}dx $$

note that the cubic polynomial in the denominator has one real root $r= -0.5437$ and a pair of complex roots, which allows the factorization

$$x^3-x^2+x+1= (x-r)(x^2+(r-1)x-1/r) $$

Then, decompose the integrand as \begin{align} \frac{x^2+1}{x^3-x^2+x+1}=\frac{A}{x-r}+\frac{Bx+C}{x^2+(r-1)x-1/r} \end{align}

and integrate accordingly.

2. For the integral $$ \int\frac{1}{x^2e^{1/x}-x}dx \overset{x\to\frac1x}=-\int \frac1{e^x-x}dx =-\sum_{k\ge0} \int x^ke^{-(k+1)x}dx $$

where each term can be integrated separately.