### Video Transcript

Evaluate the definite integral
between the limits of one and root three of negative one over one plus π₯ squared
with respect to π₯.

Here, we have a definite integral
with limits of one and a square root of three. This means weβre going to need to
use the second part of the fundamental theorem of calculus to evaluate it. This tells us that if π is a real
value function on some closed interval π to π and capital πΉ is an antiderivative
of π in that closed interval such that capital πΉ prime of π₯ is equal to π of
π₯. Then if π is Riemann integrable on
the closed interval, then we can say that the definite integral between π and π of
π of π₯ is equal to capital πΉ of π minus capital πΉ of π. Essentially, we can evaluate the
integral here by finding the antiderivative of this function negative one over one
plus π₯ squared and evaluating it between root three and one.

Now, the integral of negative one
over one plus π₯ squared is not particularly nice. But actually we spot that we can
take out any constant factors and focus on the integral itself. So here, we take our constant
factor of negative one. And weβre now looking to evaluate
the negative integral of one over one plus π₯ squared between one and root
three. Then we spot that we have the
standard result for the derivative of the inverse tan of π₯ over π. Itβs π over π squared plus π₯
squared. And this, of course, means that the
antiderivative of π over π squared plus π₯ squared must be the inverse tan of π₯
over π.

Now, if we compare the function π
over π squared plus π₯ squared with our function one over one plus π₯ squared, we
can see that π is equal to one. So if we say that π of π₯ is equal
to one over one plus π₯ squared, then the antiderivative capital πΉ of π₯ must be
the inverse tan of π₯ over one, which is simply the inverse tan of π₯. By the second part of our theorem
then, we can say that the definite integral between one and root three of one over
one plus π₯ squared with respect to π₯ is equal to the inverse tan of root three
minus the inverse tan of one. And of course, we took out that
constant of negative one at the start.

We then know that the inverse tan
of root three is π by three, and the inverse tan of one is π by four. Weβre going to find the difference
between these two fractions by using a common denominator. We multiply the numerator and
denominator of our first fraction by four and the numerator and denominator of our
second fraction by three. And weβre looking to work out
negative four π over 12 minus three π over 12. Four π by 12 minus three π by 12
is π by 12. So our answer here is negative π
by 12. Now, we have just found the general
result for the indefinite integral of π over π squared plus π₯ squared with
respect π₯ for real constants π. Itβs the inverse tan of π₯ over π
plus some constant of integration π. And rather than jumping straight
into evaluating capital πΉ of π minus capital πΉ of π, we could, of course, have
included this extra step using those square brackets.