Question Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ 1/(π‘₯(π‘₯ + 2)) dπ‘₯.

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Video Transcript

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

We’ve been told to evaluate this integral using partial fractions. Remember, using partial fractions makes the fraction much simpler and therefore much easier to integrate. We’re already dealing with a fully factorized denominator, and so we separate one over π‘₯ times π‘₯ plus two into two separate fractions. Let’s call them 𝐴 over π‘₯ plus 𝐡 over π‘₯ plus two, where 𝐴 and 𝐡 are real constants. Now, let’s look at what we need to do to the right-hand side to make it look like the left. Well, to create a common denominator, we multiply each denominator and, in fact, the numerator by the denominator of the other fraction. So, we’re going to multiply 𝐴 over π‘₯ by π‘₯ plus two and 𝐡 over π‘₯ plus two times π‘₯.

That gives us 𝐴 times π‘₯ plus two over π‘₯ times π‘₯ plus two plus 𝐡 times π‘₯ over π‘₯ times π‘₯ plus two. And now, since the denominators are the same, we can simply add the numerators. So, we get 𝐴 times π‘₯ plus two plus 𝐡π‘₯ over π‘₯ times π‘₯ plus two. And of course, this is equal to the original fraction, one over π‘₯ times π‘₯ plus two. Notice that their denominators are now the same. And we can therefore say that for these two fractions to be equal, their numerators must be equal. That is, one is equal to 𝐴 times π‘₯ plus two plus 𝐡π‘₯. Our job is to find the values of the constants 𝐴 and 𝐡, and there are a couple of ways we can do this. We could distribute the parentheses and equate coefficients. That would give us a pair of simultaneous equations, which we could solve for 𝐴 and 𝐡.

Alternatively, we substitute the roots or the zeroes of the equation 𝑦 equals π‘₯ times π‘₯ plus two. That is, we let π‘₯ be equal to negative two and π‘₯ be equal to zero. And we’ll see why this method could be quite nice in a moment. When we let π‘₯ be equal to negative two, we get one equals 𝐴 times negative two plus two plus 𝐡 times negative two. Well, negative two plus two is zero. So, we get one equals 𝐡 times negative two. And then, we divide through by negative two. And we see that 𝐡 is equal to negative one-half. Notice how by letting π‘₯ be equal to negative two, the expression 𝐴 times π‘₯ plus two became zero, leaving us an equation purely in terms of 𝐡. Next, we let π‘₯ be equal to zero. That’s one equals 𝐴 times zero plus two plus 𝐡 times zero. This time, we get 𝐡 times zero being equal to zero.

And so, we’re left with one equals 𝐴 times zero plus two or one equals two 𝐴. We solve for 𝐴 by dividing through by two to give us 𝐴 equals one-half. We now go back to the original identity, where we replace 𝐴 with one-half and 𝐡 with negative one-half. And so, to find the integral of one over π‘₯ times π‘₯ plus two with respect to π‘₯, we need to integrate one over two π‘₯ minus one over two times π‘₯ plus two with respect to π‘₯. Now remember, the integral of the sum or difference of two functions is equal to the sum or difference of the integrals of each respective function. So, we can write this as the integral of one over two π‘₯ with respect to π‘₯ minus the integral of one over two times π‘₯ plus two with respect to π‘₯. We can then take out any constant factors.

Here, we can take out a factor of one-half from our first integral and a factor of one-half from the second. And we’re now ready to integrate. The integral of one over π‘₯ with respect to π‘₯ is the natural log of the absolute value of π‘₯. Similarly, the integral of one over π‘₯ plus two with respect to π‘₯ is the natural log of the absolute value of π‘₯ plus two. And since we’re dealing with an indefinite integral, we need to add that constant of integration 𝑐. Now, we’re going to factor by one-half. Well, we get half times the natural log of the absolute value of π‘₯ minus the natural log of the absolute value of π‘₯ plus two. And then, we recall one of our laws of logs. This says that the log of 𝐴 minus the log of 𝐡 is equal to the log of 𝐴 divided by 𝐡.

And so, the natural log of the absolute value of π‘₯ minus the natural log of the absolute value of π‘₯ plus two is equal to the natural log of the absolute value of π‘₯ over the absolute value of π‘₯ plus two. But we also know that 𝐴 times the log of 𝐡 can be written as the log of 𝐡 to the power of 𝐴. So, we can write this as the natural log of the absolute value of π‘₯ over the absolute value of π‘₯ plus two to the power of one-half plus that constant of integration 𝑐. But the power of one-half is a square root. And so, we found that the indefinite integral of one over π‘₯ times π‘₯ plus two with respect to π‘₯ is the natural log of the square root of the absolute value of π‘₯ over the absolute value of π‘₯ plus two plus a constant of integration 𝑐.

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