Question

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

Answer 1

Answer:

INtegral converges to 1

Step-by-step explanation:

Given is an improper integral and limits are x=0 to infinity

f(x) = $\frac{1}{(x+1)^2}$

is to be integrated

f(x) is defined for the values from 0 to infinity since only at -1 f becomes undefined.

We have

[tex]\int \frac{1}{(x+1)^2}dx=-\frac{1}{(x+1)[tex]

Substitute limits

When we substitute infinity this becomes 0

Hence integral value is = 0-(-1/1)=1

INtegral converges to 1