### Video Transcript

Determine whether the integral from negative one to two of π₯ divided by π₯ plus one all squared with respect to π₯ is convergent or divergent.

In this question, weβre given a definite integral. And we need to determine the convergence or divergence of this definite integral. And the first question we should ask ourself is, is our integral improper? Remember, there are two different ways for an integral to be improper. First, we need to check if either of the limits of integration are positive or negative β. In this case, we can see this is not true; both of our limits of integration are finite.

The second thing we need to check is, is our integrand continuous across the entire interval of integration? In our case, the lower limit of integration is negative one and the upper limit of integration is two. So, our interval of integration is the closed interval from negative one to two. So, we need to check if our integrand is continuous on this closed interval.

And to do this, we need to notice our integrand is a rational function. And rational functions are continuous across their entire domain. And we know the only place a rational function wonβt be defined is when its denominator is equal to zero. So, we just need to solve the denominator of our integrand being equal to zero. In this case, thatβs when π₯ is equal to negative one. So, our integrand is continuous for all real numbers except when π₯ is equal to negative one. And we can see this is the lower limit of integration. So, this is an improper integral where our integrand is continuous everywhere except at the lower limit of integration.

So, to answer this question, weβre going to need to recall how we evaluate improper integrals of this form. We recall if π is continuous for all values of π₯ greater than a constant π and less than or equal to some constant π and it has a discontinuity at π₯ is equal to π, then the integral from π to π of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the right of the integral from π‘ to π of π of π₯ with respect to π₯ provided this limit exists and is finite.

In other words, we can turn a question about evaluating an improper integral into a question about the convergence or divergence of a limit. And we can see this is exactly the type of integral we have in our question. First, weβve already shown that our integrand is not defined when π₯ is equal to negative one, so itβs not continuous when π₯ is equal to negative one. And we also know itβs continuous for all values of π₯ bigger than negative one and less than or equal to two. This means we can instead try to evaluate our integral by instead evaluating this limit. And this is exactly what the question means when it asks is our integral convergent or divergent. It means, is this limit convergent or divergent?

So, to answer this question, we need to start by setting up our limit. We set π equal to negative one, π equal to two, and π of π₯ to be the integrand given to us in the question. This gives us the integral from negative one to two of π₯ divided by π₯ plus one all squared with respect to π₯ is equal to the limit as π‘ approaches negative one from the right of the integral from π‘ to two of π₯ divided by π₯ plus one all squared with respect to π₯.

So, we need to determine the convergence or divergence of this limit. And to do this, thereβs something worth pointing out. Our values of π‘ are approaching negative one from the right. This means our values of π‘ are always bigger than negative one. Theyβre never equal to negative one. And this gives us a useful piece of information. In this definite integral, our integrand is now continuous across the entire interval of integration. This is now a proper integral, so we can use any of our tools to help us evaluate this integral.

So, the next thing we need to do is evaluate this definite integral. And in fact, weβve seen a lot of different methods of evaluating the integral of rational functions. For example, to do this integral, we could use partial fractions. And this would work. We could evaluate our integral using this method. However, an easier method is to use substitution. Weβre going to use the substitution π’ is equal to π₯ plus one. To evaluate this integral by using substitution, weβre going to need to find an expression for dπ’ by dπ₯. So, weβre going to need to differentiate both sides of this expression with respect to π₯. We get dπ’ by dπ₯ is equal to one.

And we know dπ’ by dπ₯ is not a fraction. However, when weβre using integration by substitution, it can help to treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials dπ’ is equal to dπ₯. Weβre not done yet though. Remember, this is a definite integral weβre trying to evaluate by substitution, so we need to find the new limits of integration. To do this, we need to substitute our limits of integration into our expression for π’.

Weβll start by finding the new upper limits of integration. We need to substitute π₯ is equal to two into our expression for π’. We get π’ is equal to two plus one, which of course simplifies to give us three. Weβll do the same for our lower limit of integration. We substitute π₯ is equal to π‘ into our expression for π’. We get π’ is equal to π‘ plus one.

Weβre now ready to start evaluating our integral by using our substitution. First, weβll update the limits of integration. Our new lower limit of integration is π‘ plus one, and the new upper limit of integration is three. Next, in the denominator of our integrand, we can replace π₯ plus one with π’. This gives us a new denominator in our integrand of π’ squared. Next, by using our statement in terms of differentials, we know dπ’ is equal to dπ₯. So, instead of integrating with respect to π₯, we can just integrate with respect to π’. However, we see we still need to change our numerator. Our old numerator was π₯, so we need to find an expression for π₯ in terms of π’.

And we can do this by rearranging our expression for π’. We need to subtract one from both sides of the equation. This gives us that π₯ will be equal to π’ minus one. So, we replace the π₯ in the numerator of our integrand with π’ minus one. This gives us the limit as π‘ approaches negative one from the right of the integral from π‘ plus one to three of π’ minus one all divided by π’ squared with respect to π’.

Weβre now almost in a stage where we can directly evaluate this integral. Weβre just going to divide both terms in our numerator separately by π’ squared. Doing this means we would rewrite our integrand as π’ divided by π’ squared minus one over π’ squared. Now, this is almost in a form we can integrate directly. We want to cancel the shared factor of π’ in our numerator and our denominator of the first term. But to do this, we do need to be slightly careful. We need to make sure our values of π’ are never equal to zero.

And in fact, we can see this is indeed the case. First, our values of π‘ are approaching negative one from the right. So, in particular our values of π‘ must be bigger than negative one. And our values of π’ will be the interval of integration. In other words, in this integral, our values of π’ are ranging from π‘ plus one to three. And to see that our values of π’ are never equal to zero, we can add one to both sides of our inequality. We see that π‘ plus one is always bigger than zero. So, our values of π’ range from a number bigger than zero to three. So, π’ is never equal to zero. And this means in the first term in our integrand, we can cancel the shared factor of π’ in the numerator and our denominator. It wonβt change the value of our integral.

So, our integrand is now one over π’ minus one over π’ squared. And we can evaluate this integral directly. We want to evaluate our integral term by term. First, the integral of one over π’ with respect to π’ is the natural logarithm of the absolute value of π’. To help us evaluate the integral of our second term, it can help to rewrite this as π’ to the power of negative two. Then, we can just do this by using the power rule for integration.

We add one to our exponent of π’ giving us a new exponent of negative one and then divide by this new exponent of negative one. So, weβre subtracting π’ to the power of negative one divided by negative one. And of course, we can simplify this. This just simplifies to give us π’ to the power of negative one, which is one over π’.

And finally, we still need to include the limits of integration. This means we need to determine whether the following limit is convergent or divergent. To do this, we need to evaluate our antiderivative at the limits of integration. Evaluating our antiderivative at the limits of integration, we now have the limit as π‘ approaches negative one from the right of the natural logarithm of the absolute value of three plus one-third minus the natural logarithm of the absolute value of π‘ plus one minus one divided by π‘ plus one.

And this is a complicated-looking integral. However, we can evaluate this limit term by term directly. First, we see our limit is as π‘ is approaching negative one from the right. Our first two terms, the natural logarithm of the absolute value of three and one-third, are constant. They donβt vary as π‘ varies. So, their limit will just be equal to themselves. But letβs then see what happens to our third term. We see weβre taking the natural logarithm of the absolute value of π‘ plus one.

Remember, if π‘ is approaching negative one from the right, then we must have that π‘ plus one is approaching zero from the right. And this is where we get a small problem. If π‘ plus one is approaching zero from the right, we can look at the limits of our final two terms. As π‘ plus one approaches zero from the right, negative the natural logarithm of the absolute value of π‘ plus one will be approaching β. This is because weβre taking the natural logarithm of a very small positive number, which is a very large negative number, and then multiplying it by negative one.

However, in our fourth term, the opposite happens. We have one divided by a very small positive number and then we multiply it by negative one. So, our fourth term is going to limit to give us negative β. So, it appears we have an indeterminate form in our limit. We have β minus β, so we need to evaluate this limit using a different method. However, we know that rational functions will dominate the natural logarithm function. You can show this by looking at the limits of their quotients or by using something like LβHopitalβs rule.

This means that our limit will diverge. In fact, it will be equal to negative β. But remember, saying that a limit is equal to negative β is another way of saying that it diverges. Therefore, we were able to show the integral from negative one to two of π₯ divided by π₯ plus one all squared with respect to π₯ is divergent.