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

Evaluate โˆซ d๐‘ฅ/(๐‘ฅยฒ โˆ’ ๐‘ฅ + 1).

02:14

Video Transcript

Evaluate the indefinite integral of one over ๐‘ฅ squared minus ๐‘ฅ plus one with respect to ๐‘ฅ.

Now, this isnโ€™t a nice function to integrate at all. So weโ€™re going to need to do something a little bit clever. Itโ€™s certainly not the product of two functions. So weโ€™re not going to use integration by parts. But if we do something special to the denominator, we can actually use integration by substitution. Weโ€™re going to complete the square of the denominator of the expression ๐‘ฅ squared minus ๐‘ฅ plus one. Remember, we halve the coefficient of ๐‘ฅ. Here, thatโ€™s negative one, so half of that is negative one-half. We, therefore, have ๐‘ฅ 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 ๐‘ฅ minus a half all squared plus three quarters. And now, this is the integral that weโ€™re looking to evaluate.

Next, we need to spot that we know that indefinite integral of ๐‘Ž over ๐‘Ž squared plus ๐‘ฅ squared. Itโ€™s the inverse tan of ๐‘ฅ over ๐‘Ž. So to ensure that our function looks a little like this, weโ€™re going to perform a substitution. Weโ€™re going to let ๐‘ฅ minus a half be equal to ๐‘ข. Then this part will be ๐‘ข squared. The derivative of ๐‘ฅ minus one-half is one. So d๐‘ข by d๐‘ฅ equals one, which means that d๐‘ข is equal to d๐‘ฅ. So we can replace d๐‘ฅ with d๐‘ข and ๐‘ฅ minus a half with ๐‘ข. And we see that weโ€™re actually looking to find the indefinite integral of one over ๐‘ข squared plus three-quarters.

Now, this still doesnโ€™t quite look like what weโ€™re after. We need it to be ๐‘Ž squared on the denominator. Well, three-quarters is the same as the square root of three quarters squared. So ๐‘Ž 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 ๐‘ข over the square root of three quarters plus ๐‘. One divided by the square root of three quarters is two root three over three. And then we go back to our substitution ๐‘ข equals ๐‘ฅ minus a half. And we replace that in our result. And finally, we distribute our parentheses. The indefinite integral of one over ๐‘ฅ squared minus ๐‘ฅ plus one with respect to ๐‘ฅ is two root three over three times the inverse tan of root three over three times two ๐‘ฅ minus one plus the constant of integration ๐‘.

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