Question Video: Integration of Rational Functions by Partial Fractions | Nagwa Question Video: Integration of Rational Functions by Partial Fractions | Nagwa

Question Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate β«(π β΄ + 81)/(π (π Β² + 9)Β²) dπ .

09:11

Video Transcript

Use partial fractions to evaluate the indefinite integral of π  to the fourth power plus 81 over π  times π  squared plus nine squared with respect to π .

Letβs recall what we actually mean when we write a rational expression in partial fraction form. Essentially, weβre going to reverse the process for adding fractions to write π  to the fourth power plus 81 over π  times π  squared plus nine all squared as the sum of some simpler rational functions.

Now, there are several steps that we need to take to do so. The first thing weβre going to check is that the order of the expression on our denominator is greater than that of our numerator. The order or the degree of the polynomial on our denominator is five, since distributing the parentheses would give us the highest exponent of π  as five, whereas the order of our numerator is four. Since the order of the polynomial on the denominator is higher than that of the numerator, we donβt need to perform any polynomial long division. And we can move on to our second step.

Now, our second step is to fully factor the expression on the denominator. But this is done for us. And so weβre going to begin the next step. The first thing we do is we look for any nonrepeated factors in our denominator. The repeated factor here is π  squared plus nine because itβs being squared. π , however, is a nonrepeated factor, and in fact itβs a linear factor.

And so our first rational function is π΄, which is some constant, over π . We then look at our repeated factor. Since itβs being squared, weβre going to have two rational functions in increasing exponents of π  squared plus nine. So their respective denominators are π  squared plus nine and π  squared plus nine squared.

Remember though, these denominators are not linear; theyβre quadratic. And so their numerators have to reflect this. We write these as π΅π  plus πΆ plus π·π  plus πΈ, where π΅, πΆ, π·, and πΈ are constants. Our next job is to work out the value of π΄, π΅, πΆ, π·, and πΈ. And so what we do now is we add these three rational expressions. And we do so by creating a common denominator. That common denominator will match the denominator of our original rational function.

And so we see that, for our first fraction, weβll need to multiply its denominator by π  squared plus nine squared to achieve this. Of course, to create an equivalent fraction, we need to do the same to the numerator, giving us π΄ times π  squared plus nine squared over π  times π  squared plus nine squared. For our second fraction, weβll need to multiply the numerator and denominator by π  times π  squared plus nine. This gives us a numerator of π΅π  plus πΆ times π  times π  squared plus nine. Finally, weβll multiply our third fraction by π  only. And this gives us a numerator of π  times π·π  plus πΈ.

Now, since the denominators of our three fractions are equal, we simply add their numerators as shown. Now, of course, this is still equal to our earlier fraction. So we see that since the denominator of these two fractions is equal, for the fractions themselves to be equal, the numerators must also be equal. That is, π  to the fourth power plus 81 must be equal to π΄ times π  squared plus nine squared plus π΅π  plus πΆ times π  times π  squared plus nine plus π  times π·π  plus πΈ.

Now, to find the values of our constants, we have a couple of options. One method is to choose values of π . And weβre allowed to do this because, actually, we could use an identity symbol here. This expression is true for all values of π . The other method is to distribute our parentheses on the right-hand side and equate coefficients.

Now, there are very few values of π  that will help us to eliminate any terms. So weβre going to use the second method and equate coefficients. Distributing the parentheses, π  squared plus nine squared gives us π  to the fourth power plus 18π  squared plus 81. Of course, this is all being multiplied by π΄. And so we can write this as π΄π  to the fourth power plus 18π΄π  squared plus 81π΄. Then when we distribute π  times π  squared plus nine, we get π  cubed plus nine π . We can then distribute again, and we get π΅π  to the fourth power plus nine π΅π  squared plus πΆπ  cubed plus nine πΆπ . Letβs distribute our final set of parentheses. When we do, we get π·π  squared plus πΈπ .

Weβre now going to equate coefficients of π . Letβs begin by comparing the coefficients of π  to the fourth power on both sides. On the left-hand side, we simply have π  to the fourth power. So its coefficient is one. On our right-hand side, we have π΄π  to the fourth power plus π΅π  to the fourth power. So our coefficient is π΄ plus π΅. And we formed an equation; that is, one equals π΄ plus π΅.

Weβll now compare coefficients of π  cubed. On the left, thatβs zero. And on the right, we only have πΆ. And so weβve established that πΆ is equal to zero. Weβll now look at coefficients of π  squared. Once again, on the left, thatβs zero. And on the right, we have 18π΄, we have nine π΅, and we have π·. Next, itβs π  to the power of one or simply π . On the left-hand side, thatβs zero. And on the right, itβs nine πΆ plus πΈ. But of course, remember, we said πΆ is equal to zero. So this becomes zero equals nine times zero plus πΈ or zero equals zero plus πΈ, meaning that πΈ itself must also be zero.

Now, weβll consider coefficients of π  to the power of zero. Remember, π  to the power of zero is one. So weβre really looking at constant terms. On the left-hand side, itβs 81. And on the right-hand side, we have 81π΄. So, 81 equals 81π΄. And dividing through by 81 gives us π΄ equals one. We know the value of πΆ, πΈ, and π΄. So letβs go back and work out π΅ and π·. We saw that π΄ was equal to one. So our first equation becomes one equals one plus π΅, meaning π΅ must be equal to zero. Then letting π΄ be equal to one and π΅ be equal to zero in our third equation, and we get zero equals 18 plus zero plus π·, meaning π· is equal to negative 18.

Weβve now done enough to express our original function in partial fraction form. By replacing π΄, π΅, πΆ, π·, and πΈ with their respective values, we find π  to the fourth power plus 81 over π  times π  squared plus nine squared can be written as one over π  minus 18π  over π  squared plus nine squared.

Now, of course, weβre not finished. We were looking to integrate this. So weβre going to integrate one over π  minus 18π  over π  squared plus nine squared with respect to π . Letβs clear some space. When integrating the sum or difference of two functions, we can find the sum or difference of the integrals of their respective functions. This essentially means we can integrate term by term. And this is great because the integral of one over π  with respect to π  is quite straightforward. Itβs the natural log of the absolute value of π  plus some constant of integration π΄. But what about our second integral?

Well, the key here is spotting that the derivative of part of the denominator is some multiple of our numerator. That tells us we can use integration by substitution to evaluate this integral. We let our new variable π’ be equal to the inner part of the composite function on the denominator. π’ is π  squared plus nine. We then differentiate this with respect to π . So dπ’ by dπ  is two π .

Now, of course, this isnβt a fraction. But in integration by substitution, we treat it a little like one. And we can write this equivalently as dπ’ equals two π  dπ . Notice the numerator of our fraction is 18π . And if we multiply through dπ’ equals two π  dπ  by nine, we get the expression nine dπ’ on the left and 18π  dπ  on the right, meaning we can now replace 18π  dπ  with nine dπ’. And of course we can replace π  squared plus nine with π’.

And so weβre now going to integrate nine over π’ squared with respect to π’. One over π’ squared is the same as π’ to the power of negative two. So we can write our integrand as nine π’ to the power of negative two. When we integrate an expression of this form, we add one to the exponent and divide by that new value. So thatβs nine π’ to the power of negative one divided by negative one, which is negative nine over π’.

Remember, we have that constant of integration π΅ since weβre performing an indefinite integral. Now, we were actually integrating with respect to π . So letβs go back and replace π’ with our substitution, giving us negative nine over π  squared plus nine plus our constant of integration π΅. Meaning that when we integrate π  to the fourth power plus 81 over π  times π  squared plus nine squared with respect to π , we get the natural log of the absolute value of π  plus π΄ minus negative nine over π  squared plus nine plus π΅, which can be simplified further by distributing the parentheses and collecting together our constants to form one new constant πΎ. And we have the natural log of the absolute value of π  plus nine over π  squared plus nine plus πΎ.

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