# Video: Finding the Integration of a Rational Function by Completing the Square

Evaluate ∫ d𝑥/(𝑥² − 𝑥 + 1).

02:14

### Video Transcript

Evaluate the indefinite integral of one over 𝑥 squared minus 𝑥 plus one with respect to 𝑥.

Now, this isn’t a nice function to integrate at all. So we’re going to need to do something a little bit clever. It’s certainly not the product of two functions. So we’re not going to use integration by parts. But if we do something special to the denominator, we can actually use integration by substitution. We’re going to complete the square of the denominator of the expression 𝑥 squared minus 𝑥 plus one. Remember, we halve the coefficient of 𝑥. Here, that’s negative one, so half of that is negative one-half. We, therefore, have 𝑥 minus a half all squared in the brackets. Negative a half squared is one-quarter, so we subtract that one quarter. And we see that our expression is equivalent to 𝑥 minus a half all squared plus three quarters. And now, this is the integral that we’re looking to evaluate.

Next, we need to spot that we know that indefinite integral of 𝑎 over 𝑎 squared plus 𝑥 squared. It’s the inverse tan of 𝑥 over 𝑎. So to ensure that our function looks a little like this, we’re going to perform a substitution. We’re going to let 𝑥 minus a half be equal to 𝑢. Then this part will be 𝑢 squared. The derivative of 𝑥 minus one-half is one. So d𝑢 by d𝑥 equals one, which means that d𝑢 is equal to d𝑥. So we can replace d𝑥 with d𝑢 and 𝑥 minus a half with 𝑢. And we see that we’re actually looking to find the indefinite integral of one over 𝑢 squared plus three-quarters.

Now, this still doesn’t quite look like what we’re after. We need it to be 𝑎 squared on the denominator. Well, three-quarters is the same as the square root of three quarters squared. So 𝑎 here is equal to the square root of three-quarters. And of course, since the numerates of our fraction is one and not the square root of three-quarters, the integral is one divided by the square root of three-quarters times the inverse tan of 𝑢 over the square root of three quarters plus 𝑐. One divided by the square root of three quarters is two root three over three. And then we go back to our substitution 𝑢 equals 𝑥 minus a half. And we replace that in our result. And finally, we distribute our parentheses. The indefinite integral of one over 𝑥 squared minus 𝑥 plus one with respect to 𝑥 is two root three over three times the inverse tan of root three over three times two 𝑥 minus one plus the constant of integration 𝑐.