Answer:
INtegral converges to 1
Step-by-step explanation:
Given is an improper integral and limits are x=0 to infinity
f(x) =
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