 Hello, and welcome to this screencast. In this screencast, we will evaluate an example of an improper integral with an unbounded limit. We will evaluate the integral from 1 to infinity of 2x divided by the quantity x plus 1 squared, and we have a graph of the function on the right. We call this an example of an improper integral because one of the limits of integration is unbounded, and this means that the area under the curve that we would like to calculate extends to the right without bound. When we look at the behavior of the function as x increases, we see that there is a horizontal asymptote in the graph. This means that it's possible, but not guaranteed, that the area under the curve is finite, and so it is possible that the integral may converge to a finite value. Our first step in evaluating an improper integral is to replace the unbounded limit with a variable representing a finite value, and we'll use the variable b. And then we'll let b approach infinity, and we can now evaluate the new integral using the fundamental theorem of calculus. Our next step is to find an anti-derivative for our integrand. We'll need to use u substitution to help us with this, so we'll do that over here on the right. We'll let u be equal to x squared plus 1 because we see its derivative, 2x, in the numerator of the integrand. Now we can write our indefinite integral in terms of our new variable u, and the result is the integral of 1 over u squared du. To make this new integral easier to evaluate, we can rewrite the integrand as u to the negative 2 power, so we can use the power rule for integration. The anti-derivative of u to the negative 2 is negative u to the negative 1 power plus c, which we can rewrite as negative 1 over u plus c. Next we can put this back in terms of the original variable x to get negative 1 over x squared plus 1 plus c. We're going to use this result of integration to go back to evaluating our improper integral, which we can now write as the limit as b approaches infinity of negative 1 over x squared plus 1 from 1 to b. When we evaluate this at b and at 1 and subtract, the result is the limit as b approaches infinity of negative 1 over b squared plus 1 minus negative 1 over 1 squared plus 1, which we can simplify to the limit as b approaches infinity of negative 1 over b squared plus 1 plus 1 half. To evaluate this limit, we want to consider what happens to this first quantity as b approaches infinity. As b increases, b squared also increases, and the whole denominator b squared plus 1 will increase without bound. This means the term negative 1 divided by b squared plus 1 will approach 0, and then the sum of both terms will approach 1 half. And that's our final result. So to summarize what we found here, we say that the integral converges to 1 half, and this means that the area under the curve is indeed finite, even though the interval we're considering is unbounded. Thanks for watching.