Determine whether the improper integral $\int_{0}^{\infty}\frac{x^3}{1+x^4}\,dx$ exists
Solution 1:
Note that $\frac{x^3}{1+x^4}=\frac1x-\frac1{x(1+x^4)}$. Now $\int_1^\infty\frac1xdx$ is divergent, while $\int_1^\infty\frac1{x(1+x^4)}dx<\int_1^\infty\frac1{x^5}dx$ is convergent, so $\int_1^\infty\frac{x^3}{1+x^4}dx$ cannot be convergent.
Solution 2:
We have $$\frac{x^3}{1+x^4} \ge \frac{1}{2x}$$ for $x\ge1$ so the integral on $[1,\infty)$ diverges.