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: 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 $n\ne\frac1{13}$
A: 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 $n=1/13$
A: Hint: note $\ln (b^3+1) \sim \ln b^3=3\ln b$ and $\ln (13b+1) \sim \ln 13b=\ln 13+\ln b$ for $b\to+\infty$.
Hence: $$\lim_\limits{b\to +\infty} [\frac{n}{3} \ln (b^3+1)-\frac{1}{13}\ln (13b+1)]=\lim_\limits{b\to+\infty} [\left(n-\frac{1}{13}\right) \ln b - \frac{\ln 13}{13}]=\begin{cases} -\frac{\ln 13}{13}, \ if \ n=\frac{1}{13} \\ -\infty, \ if \ n<\frac{1}{13} \\ +\infty, \ if \ n>\frac{1}{13}\end{cases}.$$
A: $\int_1^{+ \infty} \left(
    \frac{nx^2}{x^3 + 1} 
    -\frac{1}{13x + 1} 
  \right) dx
$
$\frac{nx^2}{x^3 + 1} 
    -\frac{1}{13x + 1}
=\frac{nx^2(13x+1)-(x^3 + 1)}{(x^3 + 1)(13x + 1)} 
=\frac{x^3(13n-1)+nx^2-1}{(x^3 + 1)(13x + 1)} 
$.
If $13n-1 \ne 0$,
this behaves like
$\frac1{x}$
and the integral diverges.
Therefore the only $n$
for which the integral
might converge is
$n =\frac1{13}$.
For this $n$,
the integrand behaves like
$\frac1{x^2}$
and the integral of this
does converge.
