Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ (π‘₯ + 4)/((π‘₯ + 6)(π‘₯ βˆ’ 1)) dπ‘₯.

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Video Transcript

Use partial fractions to evaluate the indefinite integral of π‘₯ plus four over π‘₯ plus six times π‘₯ minus one with respect to π‘₯.

Remember, we need to rewrite this integrand using partial fractions. So we begin by reversing the process we would take when adding algebraic fractions. We write it as 𝐴 over π‘₯ plus six plus 𝐡 over π‘₯ minus one. And there are two ways for us to work out the constants 𝐴 and 𝐡. They’re substitution and equating coefficients.

Let’s begin by looking at the substitution method. Let’s imagine we’re adding these fractions. We multiply the numerator and denominator of the first fraction by π‘₯ minus one and the numerator and denominator of the second fraction by π‘₯ plus six. So we see that π‘₯ plus four over π‘₯ plus six times π‘₯ minus one is equal to 𝐴 times π‘₯ minus one plus 𝐡 times π‘₯ plus six all over π‘₯ plus six times π‘₯ minus one.

Notice that the denominator of the fractions on both sides of our equation are equal. This means that, for the fractions themselves to be equal, their numerators must be. And we can say that π‘₯ plus four is equal to 𝐴 times π‘₯ minus one plus 𝐡 times π‘₯ plus six. Okay, so far, so good.

Now we want to find a way to eliminate one of the constants from this equation. Well, we see that if we set π‘₯ to be equal to one, this bit here becomes 𝐴 times one minus one, which is 𝐴 times zero, which is zero. So let’s set π‘₯ be equal to one. When we do, we see that one plus four is equal to 𝐴 times zero plus 𝐡 times one plus six, which simplifying gives us five equals seven 𝐡. And now we have an equation in 𝐡.

So we can solve this by dividing both sides by seven. And we see that 𝐡 is equal to five-sevenths. Brilliant, so let’s repeat this process to help us establish the value of 𝐴. And you might wish to pause the video for a moment and think about what substitution would eliminate 𝐡 from this equation.

If we let π‘₯ be equal to negative six, then the second term over here becomes 𝐡 times zero. So the 𝐡’s going to be eliminated. And if we let π‘₯ be equal to negative six, our entire equation becomes negative six plus four equals 𝐴 times negative six minus one plus 𝐡 times zero, which simplifies to negative two equals negative seven 𝐴. Now we have an equation in 𝐴. And dividing both sides by negative seven, we obtain 𝐴 to be equal to two-sevenths. And we’ve successfully decomposed into partial fractions.

We can say that π‘₯ plus four over π‘₯ plus six times π‘₯ minus one is equal to two over seven times π‘₯ plus six plus five over seven times π‘₯ minus one. And we’re now of course able to integrate our expression with respect to π‘₯. Remembering that the integral of the sum of functions is the same as the sum of the integrals of those respective functions. And of course, we can take out any constant factors. And we see that our integral is equal to two-sevenths of the integral of one over π‘₯ plus six dπ‘₯ plus five-sevenths of the integral of one over π‘₯ minus one dπ‘₯.

Well, the integral of one over π‘₯ plus six is the natural log of the absolute value of π‘₯ plus six. And the integral of one over π‘₯ minus one is the natural log of the absolute value of π‘₯ minus one. And of course, since this is an indefinite integral, we must add that constant 𝑐. And we’re done. The indefinite integral of π‘₯ plus four over π‘₯ plus six times π‘₯ minus one with respect to π‘₯ is two-sevenths times the natural log of the absolute value of π‘₯ plus six plus five-sevenths of the natural log of the absolute value of π‘₯ minus one plus the constant 𝑐.

We are now going to consider how we could’ve got here using the method of equating coefficients. The starting process is the same. We need to get to this stage here. We want to distribute the parentheses on the right-hand side. And when we do, we see that π‘₯ plus four is equal to 𝐴π‘₯ minus 𝐴 plus 𝐡π‘₯ plus six 𝐡.

Now this next step isn’t entirely necessary. But it can help us figure out what to do next. We collect together like terms. And we see that π‘₯ plus four is equal to 𝐴 plus 𝐡 times π‘₯ plus negative 𝐴 plus six 𝐡 or six 𝐡 minus 𝐴. And now we have two families of terms, if you will. We have π‘₯ to the power of ones, and then we have these constants. And we can say that those are π‘₯ to the power of zeros.

We want to equate coefficients for these terms. Let’s begin by equating the coefficients of π‘₯ to the power of zero, or just the constants. On the left-hand side, we have four. And on the right-hand side, we have six 𝐡 minus 𝐴. Next, we’ll equate the coefficients of π‘₯ to the power of one. The coefficient of π‘₯ on the left-hand side is one. And on the right-hand side, that’s 𝐴 plus 𝐡. So we now have a pair of simultaneous equations which we can begin to solve by first adding.

Negative 𝐴 plus 𝐴 is zero. So we see that when we add our pair of simultaneous equations, we end up with five equals seven 𝐡. And solving this equation for 𝐡, we find that 𝐡 is equal to five-sevenths. We’re gonna then substitute this value of 𝐡 into either of our original equations. I’m going to choose this one here. So one is equal to 𝐴 plus five-sevenths. Then subtracting five-sevenths from both sides, we obtain 𝐴 to be equal to two-sevenths. And the rest of the process is exactly the same.

We have our partial fractions, and we can integrate each of them. And we obtain the indefinite integral to be equal to two-sevenths of the natural log of π‘₯ plus six plus five-sevenths of the natural log of π‘₯ minus one plus 𝑐.

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