00:01
So i asked to find the definite integral of the following function, which is 9 over 9 plus x squared dx.
00:09
And this is from negative 6 to 6.
00:11
So the first thing we're going to do is find the corresponding indefinite integral.
00:16
So we'll go ahead and do that together.
00:19
So let's go ahead and do this on the next page.
00:21
We'll do this in blue.
00:24
So i'll go ahead and rewrite this.
00:26
But the first thing we're going to do is apply the multiple constant rule.
00:30
So this turns into 9 times the integral.
00:33
Of 1 over x squared plus 9 dx great now we're going to go ahead and do some substitutions so the first substitution i'm going to do is for x i'm going to say let x equal to 3 u thus i can say that d x is equal to 3 times d u and that follows that u is equal to x over 3 so this integral becomes 9 times the integral of 1 over 3 times u squared plus 1 d u so now we're going to go ahead and apply the constant multiple rule uh once again by taking out this 1 3rd so this turns into 9 times 1 3rd times the integral of 1 over u squared plus 1 u u u.
01:32
Great so we know that the integral of 1 over u square plus 1 is familiar to us it is arc tangent so this actually turns into three times the inverse tangent or arc tangent of you so now what we have to do is recall i'll do this part in black we're going to recall what our value for u is x over 3 so this turns into three times the inverse tangent of x over three.
02:10
The last thing you have to do is apply our constant of integration, and it looks something like this...