Video: Integration by Partial Fractions

Use partial fractions to evaluate the integral ∫ 3/(5π‘₯Β² βˆ’ 20) dπ‘₯.

03:31

Video Transcript

Use partial fractions to evaluate the integral of three over five π‘₯ squared minus 20 with respect to π‘₯.

Before we do anything here, we need to notice that the denominator of our fraction has a constant factor of five. And we know that we’re allowed to take out any constant factors and focus on integrating the function itself. So we can write the integral of three over five π‘₯ squared minus 20 with respect to π‘₯ as three-fifths times the integral of one over π‘₯ squared minus four with respect to π‘₯.

We’re told we need to use partial fractions to evaluate the integral. Now what this means is we need to start with a simplified answer and take it back, decomposing the expression into its initial polynomial fractions. Now we can’t do this until we fully factor the denominator. So we factor π‘₯ squared minus four. And we get π‘₯ plus two times π‘₯ minus two.

Now we break this expression apart. And we write it as the sum of some constant 𝐴 over π‘₯ plus two and some constant 𝐡 over π‘₯ minus two. Our job now is to get this expression on the right to look a little more like that on the left. And we recall that when we add fractions, we need to create a common denominator.

Here, to achieve that, we’re going to multiply the numerator and the denominator of our first fraction by π‘₯ minus two and of our second fraction by π‘₯ plus two. And so we get 𝐴 times π‘₯ minus two over π‘₯ plus two times π‘₯ minus two plus 𝐡 times π‘₯ plus two over π‘₯ minus two times π‘₯ plus two. Notice now that the denominators are equal. And so we simply add the numerators. And we get 𝐴 times π‘₯ minus two plus 𝐡 times π‘₯ plus two over π‘₯ plus two times π‘₯ minus two.

Now of course, this is equal to our original fraction one over π‘₯ plus two times π‘₯ minus two. And since the denominators are equal, we know that, for the fractions to be equal, the numerators themselves must be equal. That is, one is equal to 𝐴 times π‘₯ minus two plus 𝐡 times π‘₯ plus two.

And now we have two choices. We could distribute the parentheses on the right-hand side, equate coefficients, and create some simultaneous equations in 𝐴 and 𝐡. However, it’s actually much simpler to substitute the roots or the zeroes of π‘₯ plus two times π‘₯ minus two. That is, π‘₯ equals two and π‘₯ equals negative two. And we’ll see why this is useful in a moment.

When we let π‘₯ be equal to two, we get one equals 𝐴 times two minus two plus 𝐡 times two plus two. Now 𝐴 times two minus two is 𝐴 times zero, which is zero. So we can see that, by substituting the roots of π‘₯ plus two times π‘₯ minus two, create in a single equation in terms of one of our constants. Here it’s in terms of 𝐡.

So simplifying, we get one equals four 𝐡. And then we divide through by four. And we find that 𝐡 is equal to one-quarter. We’ll repeat this process for π‘₯ equals negative two. We get one equals 𝐴 times negative two minus two plus 𝐡 times negative two plus two. This time, 𝐡 times negative two plus two is zero. So we get one equals negative four 𝐴. And again, we divide through by four. And we get 𝐴 equals negative one-quarter.

We now replace 𝐴 with negative one-quarter and 𝐡 with a quarter. And then we see that we can replace one over π‘₯ squared minus four with these fractions. They simplify to negative one over four times π‘₯ plus two plus one over four times π‘₯ minus two. And in fact, we might even choose to take out another constant factor of one-quarter.

So we get three twentieths times the integral of negative one over π‘₯ plus two plus one over π‘₯ minus two with respect to π‘₯. Then we recall that the integral of one over π‘₯ plus some constant π‘Ž with respect to π‘₯ is the natural log of the absolute value of π‘₯ plus π‘Ž plus a constant of integration 𝐢. And so the integral of negative one over π‘₯ plus two is negative the natural of the absolute value of π‘₯ plus two. And the integral of one over π‘₯ minus two is the natural log of the absolute value of π‘₯ minus two.

We might choose to neaten this up a little bit. Remembering that if we multiply our earlier constant by some other value, we get a new constant. Let’s call that π‘˜. And so we find that the integral of three over five π‘₯ squared minus 20 with respect to π‘₯ is three twentieths. Times the natural log of the absolute value of π‘₯ minus two. Times the natural log of the absolute value of π‘₯ plus two plus π‘˜.

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