# Video: Integration by Partial Fractions

Use partial fractions to evaluate the integral β« 3/(5π₯Β² β 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 weβ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₯.

Weβ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 canβ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 π΄ over π₯ plus two and some constant π΅ 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, weβ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 π΄ times π₯ minus two over π₯ plus two times π₯ minus two plus π΅ 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 π΄ times π₯ minus two plus π΅ 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 π΄ times π₯ minus two plus π΅ 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 π΄ and π΅. However, itβ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 weβll see why this is useful in a moment.

When we let π₯ be equal to two, we get one equals π΄ times two minus two plus π΅ times two plus two. Now π΄ times two minus two is π΄ 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 itβs in terms of π΅.

So simplifying, we get one equals four π΅. And then we divide through by four. And we find that π΅ is equal to one-quarter. Weβll repeat this process for π₯ equals negative two. We get one equals π΄ times negative two minus two plus π΅ times negative two plus two. This time, π΅ times negative two plus two is zero. So we get one equals negative four π΄. And again, we divide through by four. And we get π΄ equals negative one-quarter.

We now replace π΄ with negative one-quarter and π΅ 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 with respect to π₯ is the natural log of the absolute value of π₯ plus π plus a constant of integration πΆ. 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. Letβ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.