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?
 A: 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.
A: The only problem is in $+\infty$. We have for $x\geq 1$ that $1+x^7\leq 2x^7$ so $\frac{7x^7}{1+x^7}\geq \frac 72\geq 0$ and $\int_1^{+\infty}\frac 72dt$ is divergent, so $\int_1^{+\infty}\frac{7x^7}{1+x^7}dx$ is divergent. Finally, $\int_0^{+\infty}\frac{7x^7}{1+x^7}dx$ is divergent.
A: An inproper integral will diverge if the limit of the function at infinity is not zero (as Chris pointed out, it's a different business if the limit doesn't exist). Here, 
$$
\lim_{x\to\infty}\frac{7x^7}{1+x^7}=7,
$$
so the integral diverges.
