Question: "Determine whether improper integral converges or diverges, and..."

ruigE

Answered question

2021-11-10

Determine whether improper integral converges or diverges, and find the value of each that converges.

\int_{0}^{\infty} \frac{d x}{{(x + 9)}^{2}}

Cristiano Sears

Skilled2021-11-11Added 96 answers

Step 1
The given improper integral is

\int_{0}^{\infty} \frac{d x}{{(x + 9)}^{2}}

.
Step 2
Check whether the improper integral converges or diverges:

\int_{0}^{\infty} \frac{d x}{{(x + 9)}^{2}} = \int_{0}^{\infty} {(x + 9)}^{- 2} d x

= {[\frac{{(x + 9)}^{- 2 + 1}}{- 2 + 1}]}_{0}^{\infty}

= {[- \frac{1}{(x + 9)}]}_{0}^{\infty}

= - \frac{1}{(\infty + 9)} + \frac{1}{(0 + 9)}

= 0 + \frac{1}{9}

= \frac{1}{9}

Hence, the integral value is finite. so the given improper integral converges.
The value of integral is

\frac{1}{9}

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