# Question Video: Integration of Rational Functions by Partial Fractions Mathematics • Higher Education

Use partial fractions to evaluate β« π₯Β²/(π₯ β 1)(π₯Β² + 2π₯ + 1) dπ₯.

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### Video Transcript

Use partial fractions to evaluate the indefinite integral of π₯ squared over π₯ minus one times π₯ squared plus two π₯ plus one with respect to π₯.

We begin by recalling that we can only write a fraction in its partial fractions when its denominator is fully factored. Here, we have a quadratic expression: π₯ squared plus two π₯ plus one. Weβre going to need to factor this fully before we can use partial fractions. Well, we can write π₯ squared plus two π₯ plus one as π₯ plus one times π₯ plus one, which is, of course, π₯ plus one squared. And so, weβre going to write π₯ squared over π₯ minus one times π₯ plus one squared in partial fraction form.

Notice here, we have a factor which has an exponent. We have π₯ plus one squared. This means in partial fraction form, we write π΄ over π₯ minus one plus π΅ over π₯ plus one plus πΆ over π₯ plus one squared. Our job is to make the expression on the right look a little more like the expression on the left. And so, we recall that to add algebraic fractions, we create a common denominator. To achieve this, we multiply the numerator and denominator of our first fraction by π₯ plus one squared. For our second fraction, we multiply by π₯ minus one times π₯ plus one. And for our third fraction, we multiply by π₯ minus one only.

And this means our first fraction becomes π΄ times π₯ plus one squared over π₯ minus one times π₯ plus one squared. Our second fraction is π΅ times π₯ minus one times π₯ plus one over π₯ minus one times π₯ plus one squared. And our third fraction becomes πΆ times π₯ minus one over π₯ minus one times π₯ plus one squared. And now that we have a common denominator, we can simply add the numerators. Notice that the denominators on the right- and left-hand side of our equation are now equal. And so, for these two fractions to be equal then, the numerators must themselves be equal. And so, we can say that π₯ squared equals π΄ times π₯ plus one squared plus π΅ times π₯ minus one times π₯ plus one plus πΆ times π₯ minus one.

Our job is to find the value of the constants π΄, π΅, and πΆ. And we have two ways that we can go about this. We can distribute our parentheses and equate coefficients. So, that could be quite a long-winded method. Alternatively, we can substitute the roots or the zeros of π₯ minus one times π₯ plus one squared. That is, we let π₯ be equal to negative one. And we let π₯ be equal to one. When π₯ is equal to negative one, we get negative one squared equals π΄ times zero squared plus π΅ times negative two times zero plus πΆ times negative two. And in fact, π΄ times zero squared and π΅ times negative two times zero are themselves zero. And thatβs the whole purpose of substituting the roots in. When we do, we get an equation in terms of one of our constants, this time in terms of πΆ. When we simplify, we find that one is equal to negative two πΆ. And dividing through by negative two, we find πΆ is equal to negative one-half.

Letβs repeat this process with π₯ equals one. This time, we get one squared equals π΄ times two squared plus π΅ times zero times two plus πΆ times zero. And this time, the terms involving π΅ and πΆ become zero. So, we get one equals four π΄, which means π΄ is equal to one-quarter. So, how do we find the value of π΅? Well, we can now choose any value of π₯, since we know the values of π΄ and πΆ. Letβs let π₯ be equal to zero. By also letting πΆ be equal to negative one-half and π΄ be equal to a quarter, our equation becomes zero squared equals a quarter times one squared plus π΅ times negative one times one minus a half times negative one. This gives us zero equals a quarter minus π΅ plus one-half. And the expression on the right-hand side of this equation simplifies to three-quarters minus π΅. By adding three-quarters to both sides, we find π΅ to be equal to three-quarters.

And now that we know the values of π΄, π΅, and πΆ, weβre going to rewrite our integral. A quarter over π₯ minus one can be written as one over four times π₯ minus one. Three-quarters over π₯ plus one can be written as three over four times π₯ plus one. And negative one-half over π₯ plus one squared can be written as negative one over two times π₯ plus one all squared. We then recall that the integral of the sum or difference of a number of functions is equal to the sum or difference of the integrals of each respective function. And we also know we can take any constant factors outside the integral sign. So, we can write this as a quarter times the integral of one over π₯ minus one dπ₯ plus three-quarters times the integral of one over π₯ plus one with respect to π₯ minus a half times the integral of one over π₯ plus one squared with respect to π₯.

Letβs now integrate each one in turn. The integral of one over π₯ minus one with respect to π₯ is the natural log of the absolute value of π₯ minus one. Now, this is an indefinite integral. So, there is a constant of integration. But actually, weβre going to combine each of these at the end. Then, the integral of one over π₯ plus one with respect to π₯ is the natural log of the absolute value of π₯ plus one. But how do we integrate one over π₯ plus one squared? Well, weβre going to perform a substitution. Weβre going to let π’ be equal to π₯ plus one. This means the derivative of π’ with respect to π₯, dπ’ by dπ₯, is equal to one. And whilst dπ’ by dπ₯ isnβt a fraction, weβd treat it a little like one in this process. And we say that dπ’ is equal to dπ₯.

And weβre now able to integrate one over π’ squared with respect to π’. If we consider one over π’ squared as π’ to the power of negative two, then its integral is π’ to the power of negative one divided by negative one, which is negative one over π’. Letβs replace π’ now with π₯ plus one and we add that constant of integration. And in an unsimplified form, we have a quarter times the natural log of the absolute value of π₯ minus one plus three-quarters times the natural log of the absolute value of π₯ plus one minus a half times negative one over π₯ plus one plus πΆ. We take out a common factor of a quarter in our first two terms. And then, we multiply negative a half by negative one over π₯ plus one. And we get one over two π₯ plus two.

Then, we recall one of our laws of logs. And that says that π times log of π is equal to the log of π to the power of π. And this means we can write three times the natural log of π₯ plus one as the natural log of π₯ plus one to the power of three. But thereβs another rule we can use. And that says that the log of π plus the log of π can be written as the log of π times π. And so, we can write the natural log of the absolute value of π₯ minus one plus three times the natural log of the absolute value of π₯ plus one as the natural log of the absolute value of π₯ minus one times π₯ plus one cubed.

Now, in fact, in this question, weβre required to give our constant as a π. So, the indefinite integral of π₯ squared over π₯ minus one times π₯ squared plus two π₯ plus one with respect to π₯ is a quarter times the natural log of the absolute value of π₯ minus one times π₯ plus one cubed plus one over two π₯ plus two plus π.