Prove that the following integral is divergent $$\int_0^\infty \frac{7x^7}{1+x^7}$$
Im really not sure how to even start this. Does anyone care to explain how this can be done?

The answer by Davide Giraudo has all the right elements, but I think the OP may appreciate to see all of the details spelled out. First of all, an improper integral is defined as a limit, as follows:
$$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx.$$
Next, we use the fact that $1+x^7\leq 2x^7$ when $x\geq 1$, and the  properties of the definite integral to bound the integral inside the limit:
$$\begin{align*} \int_0^N \frac{7x^7}{1+x^7} dx &=\int_0^1 \frac{7x^7}{1+x^7} dx + \int_1^N \frac{7x^7}{1+x^7} dx \\
&\geq \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7x^7}{2x^7} dx \\
&= \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7}{2} dx \\
& = \int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2}.\end{align*}$$
Hence:
$$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx\geq \lim_{N\to \infty} \left(\int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2} \right)= \infty.$$
Therefore, the improper integral diverges.
