How to apply CT to an improper integral?

Question

For this question, I'm not sure if I'm doing it right, can anyone please help me out?

Determine whether the integral is convergent or divergent.

$\int_{1}^{\infty} \frac{1}{x+3ln(x)}$

$\frac{1}{x+3ln(x)} \ge \frac{1}{x}$

$\int_{1}^{\infty} \frac{1}{x+3ln(x)} \ge \int_{1}^{\infty} \frac{1}{x} $

$\int_{1}^{\infty} \frac{1}{x} = \lim_{A\to \infty} ln(x)$ at $\infty$ and one, which means it equals inifinity when you evaluate it, making $\int_{1}^{\infty} \frac{1}{x} $ diverge and thus, that makes $\int_{1}^{\infty} \frac{1}{x+3ln(x)}$ diverge.

Answer 1

HINT

Note that

$$ \frac{1}{x+3\ln(x)}\sim \frac1x$$

indeed for $x\to \infty$

$$\frac{\frac{1}{x+3\ln(x)}}{\frac1x}\to 1$$

then use limit comparison test with $\sum \frac1x$.

Answer 2

$\frac{1}{x + 3\ln(x)} >\frac{1}{x + 3(x-1)} =\frac{1}{4x-3}>\frac{1}{4x}$ for $x > 1$. So the integral diverges.

Answer 3

Your comparison test is incorrect. This is because:

$$\frac{1}{x+3\ln(x)} \leq \frac{1}{x} \quad x \in [1, \infty)$$

Since $\ln x \geq 0 \,\forall x \in [1, \infty)$, then the denominator must be greater than or equal to $1\over x$'s denominator. Larger denominators mean smaller quantities overall, so you cannot draw any conclusion.

Instead apply some other test, such as the limit comparison test for integrals, for convergence.