### 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.