If you mean
∫[5,-∞] 1/(x^2+1) dx
then you have
arctan(x) [5,-∞]
= -π/2 - arctan(5)
�ç5 to −�‡ (1/(x^2+1))dx
∫[5,-∞] 1/(x^2+1) dx
then you have
arctan(x) [5,-∞]
= -π/2 - arctan(5)
In this case, the function f(x) = 1/(x^2+1) is positive and continuous in the interval (-∞, ∞). However, it's not decreasing. So, we cannot directly apply the integral test.
Instead, we can apply a different test called the limit comparison test. The limit comparison test states that if f(x) and g(x) are positive functions on [a, ∞), and if lim (x→∞) f(x)/g(x) = L, where L is finite and positive, then both ∫ (a to ∞) f(x) dx and ∫ (a to ∞) g(x) dx either both converge or both diverge.
In this case, let's compare the given function with g(x) = 1/x^2. Both f(x) = 1/(x^2+1) and g(x) = 1/x^2 are positive functions on [5, ∞), and as x approaches ∞, the limit of f(x)/g(x) can be calculated:
lim (x→∞) (1/(x^2+1))/(1/x^2)
= lim (x→∞) x^2/(x^2+1)
= lim (x→∞) 1/(1+1/x^2)
= 1
Since the limit is finite and positive, we can conclude that either both ∫ (5 to ∞) f(x) dx and ∫ (5 to ∞) g(x) dx converge or both diverge.
The integral ∫ (5 to ∞) g(x) dx = ∫ (5 to ∞) 1/x^2 dx is a known convergent integral that can be evaluated using basic integration techniques:
∫ (5 to ∞) 1/x^2 dx = [ -1/x ] (from 5 to ∞)
= [ -1/∞ ] - [ -1/5 ]
= 0 + 1/5
= 1/5
Therefore, since the integral ∫ (5 to ∞) g(x) dx converges to a finite value, we can conclude that the original integral ∫ (5 to ∞) f(x) dx also converges, and its value is 1/5.