# Video: Evaluating the Definite Integration of a Standard Integral

Evaluate β«_(1) ^(β3) β(1/(1 + π₯Β²)) dπ₯.

02:58

### 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 weβre 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 πΉ is an antiderivative of π in that closed interval such that capital πΉ 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 πΉ of π minus capital πΉ 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 weβre 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. Itβs 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 πΉ 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. Weβre 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 weβre 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. Itβs the inverse tan of π₯ over π plus some constant of integration π. And rather than jumping straight into evaluating capital πΉ of π minus capital πΉ of π, we could, of course, have included this extra step using those square brackets.