Determine if the improper integral is convergent or divergent, and find its value if it is convergent.
\[
\int_{0}^{\infty} 2 x e^{3 x} d x
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The integral is convergent and $\int_{0}^{\infty} 2 x e^{3 x} d x=\square$ (Type an integer or a fraction. Simplify your answer.)
B. The integral diverges.

Step 1 ：The integral is an improper integral because it has an infinite limit. To determine if it is convergent or divergent, we can use the integral test. The integral test states that if the integral of a function from a to infinity is finite, then the series is convergent. If the integral is infinite, then the series is divergent.

Step 2 ：To find the value of the integral, we can use integration by parts. The formula for integration by parts is \(\int udv = uv - \int vdu\). We can let \(u = x\) and \(dv = 2e^{3x}dx\). Then we need to find du and v.

Step 3 ：Let's find du and v. We have \(u = x\), \(dv = 2e^{3x}dx\), \(du = dx\), and \(v = \frac{2}{3}e^{3x}\).

Step 4 ：Substitute these into the integration by parts formula, we get \(\int_{0}^{\infty} 2 x e^{3 x} d x = \left[\frac{2}{3}x e^{3x}\right]_{0}^{\infty} - \int_{0}^{\infty} \frac{2}{3} e^{3x} dx\).

Step 5 ：Evaluating the integral \(\int_{0}^{\infty} \frac{2}{3} e^{3x} dx\), we get \(-\frac{2}{9}e^{3x}\) evaluated from 0 to infinity.

Step 6 ：Substituting the limits of integration, we get \(\left[\frac{2}{3}x e^{3x}\right]_{0}^{\infty} - \left[-\frac{2}{9}e^{3x}\right]_{0}^{\infty} = \infty - \left(-\frac{2}{9}\right) = \infty\).

Step 7 ：The integral is infinite at the upper limit, which means the integral is divergent.

Step 8 ：\(\boxed{\text{The integral diverges.}}\)