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

I'm not sure how to go about this.

Working
This is convergent if 
$$
  \lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx
$$
exists.
The indefinite integral is
\begin{equation*}
  \begin{aligned}
    \int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx
    &  = \int \left(\frac{nx^2}{x^3 + 1}
    \right) - \int \left(\frac{1}{13x + 1} \right)dx \\
    &= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) 
  \end{aligned}
\end{equation*}
Which gives
\begin{equation*}
  \begin{aligned}
    &\lim_{b \to + \infty}
    \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\
    &= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\
    &= \lim_{b \to + \infty}
    \left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right)
    \\ &= \lim_{b \to + \infty}
    \left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right)
    \\ &= \lim_{b \to + \infty}
    \left(
      \frac{n}{3}
      \left(
        \ln(b^3 + 1) -  \ln(2)
      \right)
      - \frac{1}{13}
      \left(
        \ln(13b + 1)
        -  \ln(14)
      \right)
    \right)
    \\ &= \lim_{b \to + \infty}
    \left(
      \frac{n}{3}
      \left(
        \ln \left(\frac{b^3 + 1}{2}\right) 
      \right)
      - \frac{1}{13}
      \left(
        \ln \left(\frac{13b + 1}{14}\right)
      \right)
    \right)
  \end{aligned}
\end{equation*}
I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( +
\infty)$. But things went pretty south.
So I'm sure there's a better approach. 

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 $n\ne\frac1{13}$
