# Video: Determining Whether the Improper Integral of a Rational Function with Discontinuity Is Convergent or Divergent Using Integration by Partial Fractions

Determine whether the integral β«_(0)^(4) 1/(π₯Β² β π₯ β 2) dπ₯ is convergent or divergent.

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### Video Transcript

Determine whether the integral from zero to four of one divided by π₯ squared minus π₯ minus two with respect to π₯ is convergent or divergent.

The question gives us a definite integral where our integrand is a rational function. We need to determine whether this integral is convergent or divergent. To check whether a definite integral is convergent or divergent, there are a few things we need to check. First, we want to check whether our integrand is continuous on the interval of integration. In our case, weβre integrating from π₯ is equal to zero to π₯ is equal to four. So, we want to check if our integrand is continuous on the closed interval from zero to four.

In our case, our integrand is a rational function. And we know that all rational functions are continuous everywhere except where their denominator is equal to zero. So, to find any points of discontinuity, we need to fully factor our denominator. We see that π₯ squared minus π₯ minus two is equal to π₯ minus two times π₯ plus one. Then, solving our denominator equal to zero, we must have one of our factors is equal to zero. This gives us π₯ is equal to two or π₯ is equal to negative one.

So, what weβve shown is, on our closed interval from zero to four, our integrand is continuous everywhere except when π₯ is equal to two. When our integrand is not continuous over our interval of integration, we call this an improper integral. Before we use our rules for improper integrals to help us evaluate our integral, weβll first use our rules for definite integrals to split this integral into two different integrals. Weβll split this at our point of discontinuity, when π₯ is equal to two.

Weβll now recall our rules for evaluating improper integrals. First, if we have a function π which is continuous on the closed interval from π to π, except at the point where π₯ is equal to π, and our function π has a discontinuity when 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 left of the integral from π to π‘ of π of π₯ with respect to π₯ as long as this limit exists. We see we can apply this to the first one of our integrals. Doing this, we get the limit as π‘ approaches two from the left of the integral from zero to π‘ of one divided by π₯ squared minus π₯ minus two with respect to π₯.

Weβll now recall a second rule for improper integrals. If a function π is continuous on the closed interval from π to π except when π₯ is equal to π and our function π has a discontinuity when 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₯. And this is, of course, only true if this limit exists. We can see that this applies to the second one of our integrals.

Then applying this, we get the limit as π‘ approaches two from the right of the integral from π‘ to four of one divided by π₯ squared minus π₯ minus two with respect to π₯. This means we now have two definite integrals where our limit is π‘. In fact, π‘ is never equal to two. So, our integrands are continuous over both of our intervals of integration. This means we can evaluate these integrals by using any of our tools for integration.

And we know a few different ways of evaluating integrals of this form. For example, we could use a π’ substitution, and this will work. However, weβre going to do this by using partial fractions. First, remember, we can rewrite our integrand as one divided by π₯ minus two times π₯ plus one. And we can see that our denominator has two unique roots. So, by using partial fractions, we can rewrite our integrand as π΄ divided by π₯ minus two plus π΅ divided by π₯ plus one for some constants π΄ and π΅.

Weβll simplify this expression by multiplying both sides through by π₯ minus two times π₯ plus one. Doing this and then simplifying our answer, we get one is equivalent to π΄ times π₯ plus one plus π΅ times π₯ minus two. And remember, this expression will be true for all values of π₯. So, we can eliminate our variable π΄ by substituting π₯ is equal to negative one. And we can eliminate our variable π΅ by substituting π₯ is equal to two.

Letβs start by substituting π₯ is equal to negative one. We get one is equal to π΅ times negative one minus two. And we can solve this equation for π΅. We get π΅ is equal to negative one-third. Weβll now substitute π₯ is equal to two to eliminate our variable π΅. This gives us one is equal to π΄ times two plus one. And we can then solve this equation for π΄. We get π΄ is equal to one-third. So, all we need to do now is substitute this value of π΄ and this value of π΅ into our expression for partial fractions.

And so, substituting in π΄ is one-third and π΅ is negative one-third, weβve shown by using partial fractions our integrand is the same as one over three times π₯ minus two minus one over three times π₯ plus one. So, letβs now use this to rewrite the integrand inside both of our limits. We write in the integrand inside both of our limits.

And clearing some space, we now have the limit as π‘ approaches two from the left of the integral from zero to one of one over three times π₯ minus two minus one over three times π₯ plus one with respect to π₯. Plus the limit as π‘ approaches two from the right of the integral from π‘ to four of one over three multiplied by π₯ minus two minus one over three multiplied by π₯ plus one with respect to π₯.

And although these are complicated-looking integrals, we could actually just evaluate this term by term directly at this point. However, thereβs one more step of simplification we can do. Both integrals have a constant factor of one-third, which we can take outside of our integral. This gives us the following expression. Now, we see we only need to integrate one over π₯ minus two and negative one over π₯ plus one. And now, we can integrate these term by term.

We recall for any real constant π, the integral of one divided by π₯ plus π with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus π plus the constant of integration. So, letβs now evaluate this integral term by term. First, the integral of one divided by π₯ minus two is the natural logarithm of the absolute value of π₯ minus two. Next, we need to subtract the integral of one divided by π₯ plus one. Thatβs the natural logarithm of the absolute value of π₯ plus one. And remember, we need to evaluate this at the limits of our integral: π₯ is equal to zero and π₯ is equal to π‘.

We can then do exactly the same to evaluate the integral inside of our other limit. We have the integral of one over π₯ minus two is the natural logarithm of the absolute value of π₯ minus two. And then, we subtract the integral of one divided by π₯ plus one. Thatβs the natural logarithm of the absolute value of π₯ plus one. At this point, we could just evaluate both of our expressions at the limit of our integral. However, weβll use our log laws and our absolute value laws to simplify our expression first.

Remember, one of our log laws tells us the difference between two logs is the same as taking the logarithm of their quotient. And the quotient of absolute values is equal to the absolute value of their quotient. So, using both of these two rules, we can simplify this expression to give us the natural logarithm of the absolute value of π₯ minus two over π₯ plus one. So, rewriting both of these terms of the natural logarithm of the absolute value of π₯ minus two over π₯ plus one. We get the following expression which we need to evaluate.

The next thing we need to do is evaluate both of these expressions at the limits of our integral. Evaluating our first expression at the limits of our integral, we get the limit as π‘ approaches two from the left of one-third times. The natural logarithm of the absolute value of π‘ minus two divided by π‘ minus one minus the natural logarithm of the absolute of value zero minus two divided by zero plus one. And we can simplify this since zero minus two divided by zero plus one is equal to negative two. And then, of course, the natural logarithm of the absolute value of negative two is just the natural logarithm of two.

So, weβve simplified our first limit to the following expression. Weβll then do something very similar to evaluate our second expression at the limits of integration. We get the limit as π‘ approaches two from the right of one-third times the natural logarithm of the absolute value of four minus two. Divided by four plus one minus the natural logarithm of the absolute value of π‘ minus two divided by π‘ plus one. And we can then simplify this. Four minus two divided by four plus one is two over five or 0.4. So, we simplified our integral into the following two limits.

Remember, for our integral to be convergent, both of these limits must be convergent. Otherwise, our integral will be divergent. Before we start evaluating our first limit, weβll take the constant factor of one-third outside of our limit. This gives us one-third times the limit as π‘ approaches two from the left of the natural logarithm of the absolute value of π‘ minus two. Divided by π‘ plus one minus the natural logarithm of two.

There are several different ways of seeing that this limit does not converge. One way is to consider what happens inside of our limit as π‘ approaches two from the left. We see that the logarithm of two remains constant as π‘ approaches two from the left. However, when we consider this for our logarithm function, we have a problem. Our logarithm function is getting closer and closer to the logarithm of zero. And itβs approaching this from the right. And we know that this is unbounded; itβs just going to keep growing and growing. So, this limit does not exist.

And since one of our limits did not converge, this means that our integral does not converge. Therefore, weβve shown the integral from zero to four of one divided by π₯ squared minus π₯ minus two with respect to π₯ is divergent.