This AP Calculus BC Integral Looks Impossible

Added: 2026-05-13

[00:00:00]Most of the AP Calculus BC takers are thinking about simple U substitution to evaluate this integral.

[00:00:06]But this integral turns into two integrals, one for logarithm, the other one for arctangent. So, we can easily check that simple U substitution is not going to be working by getting derivative on your denominator.

[00:00:17]So, since integrand is a rational function, let me just try to get derivative of your denominator with respect TO THE X.

[00:00:30]SO, DERIVATIVE OF YOUR DENOMINATOR TURNS OUT TO be 6x + 2, which is not the same as your numerator, that is 5x + 8.

[00:00:39]So, if you're using U substitution directly to this integral, it's not going to be working. So, we need to think about a way to make U substitution to work. So, one way is to represent this 5 + 8 using 6x + 2.

[00:00:51]So, represent 5x + 8 using 6 x + [applause] 2.

[00:00:58]It's a little trick. So, we can make an equation for 5x + 8 is equal to some constant A times entire 6x + 2 + B.

[00:01:07][applause] Let's get the value of A and B by distributing this A to those two terms inside of the parentheses, right?

[00:01:14]So, using this trick, so we can rewrite the right hand side as now 6 [applause] A x + 2 A + B.

[00:01:23]So, letting the 6 A as 5 because it's the coefficient of the X, right? So, 6 A is equal to 5, meaning A is equal to 5 over 6.

[00:01:36]And since 2A + B has to be the same as 8, right? And then A is 5 over 6, so 2 * 5 over 6 + B is equal to 8.

[00:01:45]So, that says 10 over 6 + B is equal to 8.

[00:01:51]So, if you're solving this, B has to be the same as, okay, 38 over 6, which is to 19 over 3.

[00:01:59]So, 5 8 5x + 8 is the same as Now, then 5/6 parentheses 6x + 2, and then plus 19 over 3.

[00:02:13]Okay, so let me just call this integral as just the I, right?

[00:02:17]Now, we can separate this integral I into two integrals. So, integral I is the same as Let me pull this 5/6 out out of your integral. So, 5/6 times integral from 0 to 1.

[00:02:31]Okay, that of 6 x + 2. Then over your denominator 3x ^ 2 + 2x + 2 dx.

[00:02:43]And then that plus also pulling this 19 over 3 out side of your integral, and integral from 0 to 1 of now 1 over the denominator 3x ^ 2 + 2x + 2, and we have dx.

[00:03:04]Okay, so this is your integral I, right?

[00:03:07]So, we have two integrals. Integral from 0 to 1 6x + 2 over 3x ^ 2 + 2x + 2 dx, and the other integral is integral from 0 to 1 1 over 3x ^ 2 + 2x + 2 dx.

[00:03:19]Let me just call this first integral as just number one.

[00:03:23]And second one as number two.

[00:03:25]The first one get you the logarithm, right? Because now the simple U substitution will be working for the first integral. So, let's just talk about the first integral one.

[00:03:35]Okay, this is integral from 0 to 1 of 6 x + 2 over 3x ^ 2 + 2x + 2 dx.

[00:03:51]So, now we can let you as your denominator, right?

[00:03:54]So, call your you as your entire denominator 3x² + 2x + 2. Then your du is the same as 6x + 2 entire thing dx.

[00:04:06]So, then we can say dx is the same as du over 6x + [applause] 2.

[00:04:14]And I'm not going to be changing this lower bound and the upper bound by eventually working on the x, right?

[00:04:23]Okay, so your integral is going to be then the same as We still have this 5 over 6, right? So, we have 5 over 6 of integral of 6 >> [applause] >> x + 2 that over u, then that of now du over 6x + 2.

[00:04:43]So, entire terms is now canceled out.

[00:04:45]Then this is going to be the same as 5 over 6 of natural log ln. And inside, let me just have this entire u 3x² + 2x + 2.

[00:04:56]By having the same lower bound and the upper bound from 0 to 1.

[00:05:01]Okay, so making a calculation for this, it is going to be then the same as Okay, this will be the same as now then 5 over 6 of ln of making a calculation then it has to be >> [applause] >> 7 over 2.

[00:05:18]So, the first integral part is going to be 5 over 6 * ln of 7 over 2.

[00:05:23]All we need to do is to work on the second integral, right? So, we need to work on this integral from 0 to 1 of 1 over 3x² + 2x + 2.

[00:05:38]Then we have dx.

[00:05:41]So, this is the part that we'll be using arc tangent, right?

[00:05:45]So, your denominator 3x² + 2x + 2 is not going to be factored out using only the integer nicely, right? So, we'll be completing the square. So, for the denominator part, 3x ^ 2 + 2x + 2.

[00:06:03]Let's work on completing the square, right? So, this is the same as then >> [applause] >> 3 { parenthesis x ^ 2 + 2/3 x + 2.

[00:06:15]And working on this part out, then it has to be the same as then 3 Let me make a bracket of x + 1/2 of 2/3, that is 1/3 ^ 2.

[00:06:26]And then that minus 1/9 close your bracket + 2.

[00:06:32]So, let's just complete this, right?

[00:06:34]Then this has to be then the same as 3 * { parenthesis x + 1/3 ^ 2. And then -1/3 + 2 is going to be then + 5/3.

[00:06:47]So, this is going to be your denominator.

[00:06:52]Okay, so then your integral has to be integral from 0 to 1 of 1 over This whole thing is your denominator. 3 { parenthesis x + 1/3 ^ 2 + 5/3 and then we have dx.

[00:07:05]Let's pull 1/3 outside of this integral, right? Then we should have 1/3 * now integral uh from 0 to 1 of 1 over just a x + 1/3 ^ 2. And then that + 5/9 then we have dx.

[00:07:31]Now, we can think about this arc tangent integral part, right? So, say if you have integral of dx over x ^ 2 + a ^ 2.

[00:07:40]This is going to be then the same as one over a times tangent inverse or tangent of x over a.

[00:07:52]Okay, so using this, let's just rewrite this integral part, right?

[00:07:57]Then what we need to work on has to be then the same as Okay, so let's take a look at it. It is going to be then pulling one over square root of a five out, right?

[00:08:09]Okay, then that times arc tangent Let me make a bracket.

[00:08:13]arc tangent of We have x plus one over three squared, so we should have now three x plus one over square root of five.

[00:08:26]Okay, then close the bracket.

[00:08:30]And the lower bound is actually from zero, upper bound is now one.

[00:08:34]Something we need to calculate, right?

[00:08:38]So making a calculation for this, then it has to be then the same as one over square root of five.

[00:08:45]Okay, then that times arc tangent So let me just make a parenthesis. arc tangent of Plugging in one, so four over square root of a five. And then minus arc tangent of one over square root of five. Okay.

[00:09:07]And I'll be using this arc tangent formula, which is about arc tangent of a minus arc tangent of b. This is the same as arc tangent of a minus b that over one plus a b.

[00:09:26]So I'll be using this.

[00:09:32]So we can treat your a as now four over square root of five and b as one over square root of five, right? So, if you're making a calculus for this, it is going to be the same as arc tangent of square root of the 5 over 3.

[00:09:49]If you rationalize this.

[00:09:51]Okay, so that is why the answer for this question, the integral I is going to be the same as Okay, so the first part is going to be now 5 over 6 times ln of 7 over 2. Then we have plus 19 over 3 times this integral, right? The second integral.

[00:10:14]So, multiplying 19 over 3 to the value, right? So, we should have the 19 over 3 times square root of 5. So, plus 19 over 3 times square root of the 5.

[00:10:28]Okay, then that times this value.

[00:10:32]Arc tangent of square root of the 5 over 3.

[00:10:37]This has to be the answer for the question.

[00:10:44]Okay, so this is the answer for the question.

[00:10:47]And remember this integral was from 2012 AP Calculus BC exam. And then we used U substitution but not directly to this integral, but we did something to make it happen. How amazing.