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.