Video Transcript
Evaluate the indefinite integral of
one over ๐ฅ squared minus ๐ฅ plus one with respect to ๐ฅ.
Now, this isnโt a nice function to
integrate at all. So weโre going to need to do
something a little bit clever. Itโs certainly not the product of
two functions. So weโre not going to use
integration by parts. But if we do something special to
the denominator, we can actually use integration by substitution. Weโre going to complete the square
of the denominator of the expression ๐ฅ squared minus ๐ฅ plus one. Remember, we halve the coefficient
of ๐ฅ. Here, thatโs negative one, so half
of that is negative one-half. We, therefore, have ๐ฅ 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 ๐ฅ minus a half all squared plus three quarters. And now, this is the integral that
weโre looking to evaluate.
Next, we need to spot that we know
that indefinite integral of ๐ over ๐ squared plus ๐ฅ squared. Itโs the inverse tan of ๐ฅ over
๐. So to ensure that our function
looks a little like this, weโre going to perform a substitution. Weโre going to let ๐ฅ minus a half
be equal to ๐ข. Then this part will be ๐ข
squared. The derivative of ๐ฅ minus one-half
is one. So d๐ข by d๐ฅ equals one, which
means that d๐ข is equal to d๐ฅ. So we can replace d๐ฅ with d๐ข and
๐ฅ minus a half with ๐ข. And we see that weโre actually
looking to find the indefinite integral of one over ๐ข squared plus
three-quarters.
Now, this still doesnโt quite look
like what weโre after. We need it to be ๐ squared on the
denominator. Well, three-quarters is the same as
the square root of three quarters squared. So ๐ 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 ๐ข 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 ๐ข equals ๐ฅ minus a half. And we replace that in our
result. And finally, we distribute our
parentheses. The indefinite integral of one over
๐ฅ squared minus ๐ฅ plus one with respect to ๐ฅ is two root three over three times
the inverse tan of root three over three times two ๐ฅ minus one plus the constant of
integration ๐.
