Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent

Compute the definite integral:
${\int }_3^{\mathrm{\infty }}\frac{4}{{(x+4)}^3}dx$
Factor out constants:
$=4{\int }_3^{\mathrm{\infty }}\frac{1}{{(x+4)}^3}dx$
For the integrand $\frac{1}{{(x+4)}^3}$, substitute u = x + 4 and du = dx.
This gives a new lower bound $u=4+3=7$ and upper bound $u=\infty$:
$=4{\int }_7^{\mathrm{\infty }}\frac{1}{u^3}du$
Apply the fundamental theorem of calculus.
The antiderivative of $\frac{1}{u^3}is-\frac{1}{2u^2}$:
$=\underset{b\to \mathrm{\infty }}{lim}(-\frac{2}{u^2}){|}_7^b$
Evaluate the antiderivative at the limits and subtract.
$(-\frac{2}{u^2}){|}_7^{\mathrm{\infty }}=(-\frac{2}{{\mathrm{\infty }}^2})-(-\frac{2}{7^2})=\frac{2}{49}$:
Answer:
$=\frac{2}{49}$