Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$

A simpler approach:

Since we know that:

$$\int_1^\infty\frac1{x^a}~\mathrm dx<\infty\iff a>1$$

It follows that if we know

$$\lim_{x\to\infty}\frac{\frac{nx^2}{x^3+1}-\frac1{13x+1}}{1/x^a}=c\ne0$$

Then the integral converges iff $a>1$.

---

For $n=\frac1{13}$, we find that using $a=2$ satisfies the limit, so it converges for $n=\frac1{13}$.

For $n\ne\frac1{13}$, we find that using $a=1$ satisfies the limit, so it diverges for $n\ne\frac1{13}$

Note that we have

$$\begin{align} \lim_{b\to \infty}\left(\frac n3\log\left(b^3+1\right)-\frac1{13}\log\left(13b+1\right)\right) &= -\frac1{13}\log(13)\\\\ &+\lim_{b\to \infty}\left(n\log(b)-\frac1{13}\log(b)\right)\\\\ &+\frac n3 \lim_{b\to \infty}\log\left(1+\frac1{b^3}\right)\\\\&-\lim_{b\to \infty}\log\left(1+\frac{1}{13b}\right)\\\\ &=-\frac1{13}\log(13)+\lim_{b\to \infty}\left((n-1/13)\log(b)\right) \end{align}$$

which converges if and only if $n=1/13$