Question Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ dπ‘₯/(π‘₯(π‘₯Β² + 1)Β²).

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Video Transcript

Use partial fractions to evaluate the indefinite integral of dπ‘₯ over π‘₯ times π‘₯ squared plus one squared.

We’re told to use partial fractions to evaluate our integral. What that means is we’re going to write one over π‘₯ times π‘₯ squared plus one all squared, as the sum of two or more slightly simpler rational functions. Now, we need to be a little bit careful here. One of the factors in our denominator is a quadratic. It’s π‘₯ squared plus one. And this itself is being squared. So because it’s being squared, because we have a repeated factor, we need both π‘₯ squared plus one and π‘₯ squared plus one squared as two of the denominators on our rational functions. But also because this factor is a quadratic, the numerator needs to be of this form, of 𝐡π‘₯ plus 𝐢 or 𝐷π‘₯ plus 𝐸.

Now here, 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 are constants that we need to work out. So how do we do that? Well, we try to make the expression on the right look like the expression on the left. And so we create a common denominator of π‘₯ times π‘₯ squared plus one all squared. To achieve this, we multiply the numerator and denominator of our first fraction by π‘₯ squared plus one all squared, of our second fraction by π‘₯ times π‘₯ squared plus one, and of our third fraction by π‘₯. Now, the denominators are equal and we can add the numerators. So the numerator becomes 𝐴 times π‘₯ squared plus one all squared plus π‘₯ times 𝐡π‘₯ plus 𝐢 times π‘₯ squared plus one plus π‘₯ times 𝐷π‘₯ plus 𝐸. And of course, this is equal to our original fraction.

And so because the denominators of both our fractions are equal, this tells us that the numerators themselves must also be equal. That is, one must be equal to 𝐴 times π‘₯ squared plus one squared plus π‘₯ times 𝐡π‘₯ plus 𝐢 times π‘₯ squared plus one plus π‘₯ times 𝐷π‘₯ plus 𝐸. Now, to find out the values of the constants, notice how the second two terms have a factor of π‘₯. And so if we let π‘₯ be equal to zero in this equation, those two terms are essentially going to disappear. In fact, we get one equals 𝐴 times zero squared plus one squared plus zero plus zero. And of course, that’s zero. And actually, this simplifies really nicely. And we find that 𝐴 is equal to one.

Let’s now replace 𝐴 with one in our earlier equation. Next, we distribute our parentheses. And I’m going to order the terms by their exponents. When we do, we get π‘₯ to the fourth power plus 𝐡π‘₯ to the fourth power plus 𝐢π‘₯ cubed plus two π‘₯ squared plus 𝐡π‘₯ squared plus 𝐷π‘₯ squared plus 𝐢π‘₯ plus 𝐸π‘₯ plus one. We’re now going to equate coefficients of π‘₯ on the left- and right-hand side. We’ll begin by comparing coefficients of π‘₯ to the power of four. And on the left-hand side, there are no terms with that power, so we write zero. And on the right, we have π‘₯ to the fourth power plus 𝐡π‘₯ to the fourth power. So the coefficients there are one plus 𝐡.

If we subtract one from both sides, we find that 𝐡 is equal to negative one. Let’s repeat this process with π‘₯ cubed. On the left, we have zero and, on the right, we have 𝐢. So actually, 𝐢 is equal to zero. Then for π‘₯ squared, we get zero equals two plus 𝐡 plus 𝐷. But of course we saw that 𝐡 was equal to negative one. And if we simplify and solve for 𝐷, we find that 𝐷 is equal to negative one, too. Finally, we compare coefficients of π‘₯ or π‘₯ to the power of one. We get zero equals 𝐢 plus 𝐸. But we know that 𝐢 is zero and so 𝐸 is also zero. We now replace 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 in our earlier fractions with these values.

Now that we’ve done so, we can clear some space and evaluate our integral. We’re going to integrate one over π‘₯ minus π‘₯ over π‘₯ squared plus one minus π‘₯ over π‘₯ squared plus one all squared with respect to π‘₯. Well, the integral of one over π‘₯ is the natural log of the absolute value of π‘₯. But what about the other two integrals? Well, let’s integrate negative π‘₯ over π‘₯ squared plus one. We’ll do this by letting 𝑒 be equal to π‘₯ squared plus one. This means d𝑒 by dπ‘₯ is equal to two π‘₯. And we can, therefore, say even though d𝑒 by dπ‘₯ isn’t a fraction, we do treat it a little like one. And a half d𝑒 equals π‘₯ dπ‘₯.

By replacing π‘₯dπ‘₯ with a half d𝑒 and π‘₯ squared plus one with 𝑒, we can see that we’re looking to integrate negative one over two 𝑒 with respect to 𝑒. Well, that’s negative one-half the natural log of the absolute value of 𝑒. And of course, if we then replace 𝑒 with π‘₯ squared plus one, we see that when we integrate this term, we get negative a half the natural log of the absolute value of π‘₯ squared plus one. We’re actually going to perform the exact same substitution for our third term. This time, though, when we do the substitution, we get the integral of negative one over two 𝑒 squared with respect to 𝑒.

Let’s write this as negative one-half 𝑒 to the power of negative two. When we integrate this, we get negative a half times 𝑒 to the power of negative one divided by negative one. That’s a half 𝑒 to the power of negative one or one over two 𝑒. And so we replace 𝑒 with π‘₯ squared plus one. And our third term integrates to one over two times π‘₯ squared plus one. Now, remember, this is an indefinite integral. So we add that constant of integration π‘˜. All that’s left to do is to distribute the parentheses in our third term. And we’ve evaluated our integral. It’s the natural log of the absolute value of π‘₯ minus a half times the natural log of the absolute value of π‘₯ squared plus one plus one over two π‘₯ squared plus two plus the constant of integration π‘˜.

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