Finding the limit of an integral.

How do I find the limit of the following:
lim(n-->infinity) of the integral 1/(3x + 1)dx from (x=1 to n)

First, evaluate the integral in terms of n.
∫ 1/(3x + 1) dx (from x=1 to n) = (1/3)ln|3x + 1| (evaluated from x=1 to n)
= (1/3)[ln(3n + 1) - ln(4)].

Thus:
lim (n-->∞) ∫ 1/(3x + 1) dx (from x=1 to n) = 1/3 * lim (n-->∞) [ln(3n + 1) - ln(4)].

Then, since ln(3n + 1) --> ∞ as n --> ∞:
1/3 * lim (n-->∞) [ln(3n + 1) - ln(4)] = ∞.

So, the required limit equals ∞ (in other words, the integral is divergent).

I hope this helps!

Indefinite integral equals to
1/3 log(1+3 x) + constant
where log(x) is the natural logarithm.
Thus definite integral from x = 1 to n equals to
1/3 log(1 + 3n) - 1/3 log(4)
log(1 + 3n) -> +infinity when n -> infinity
log(4) is a constant
thus the limit equals +infinity