 # Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent geduiwelh 2021-09-11 Answered

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent".
${\int }_{3}^{\mathrm{\infty }}\left(\frac{8}{{\left(x+4\right)}^{3}}/2\right)$

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Compute the definite integral:
${\int }_{3}^{\mathrm{\infty }}\frac{4}{{\left(x+4\right)}^{3}}dx$
Factor out constants:
$=4{\int }_{3}^{\mathrm{\infty }}\frac{1}{{\left(x+4\right)}^{3}}dx$
For the integrand $\frac{1}{{\left(x+4\right)}^{3}}$, 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 $\frac{1}{{u}^{3}}is-\frac{1}{2{u}^{2}}$:
$=\underset{b\to \mathrm{\infty }}{lim}\left(-\frac{2}{{u}^{2}}\right){|}_{7}^{b}$
Evaluate the antiderivative at the limits and subtract.
$\left(-\frac{2}{{u}^{2}}\right){|}_{7}^{\mathrm{\infty }}=\left(-\frac{2}{{\mathrm{\infty }}^{2}}\right)-\left(-\frac{2}{{7}^{2}}\right)=\frac{2}{49}$:
$=\frac{2}{49}$