00:01
Here, we're given an indefinite integral that we want to compute.
00:06
Now, the first thing i'm going to notice is that we have something inside of a square root, whose derivative in some fashion appears in the numerator.
00:18
So let's try u substitution.
00:21
So if i say that u is equal to the stuff inside the square root, that's 7ax plus bx to the 7th, du, dx will be the derivative of that.
00:33
So the derivative of 7ax with respect to x is just 7a, plus the derivative of bx to the 7th is 7b x to the 6th.
00:44
And if i factor out a 7 from that, and also i'm going to go ahead and multiply by dx to get the du is equal to 7 times a plus bx to the 6th, dx.
00:59
Now we can start to see that this part right here is this part right here.
01:06
So really, i could rewrite this whole integral as the integral of that green portion is just du over the square root of ru.
01:20
So let's rewrite this one last time to say this is the integral of u to the negative one -half power, du.
01:27
Oh, but wait, i made a mistake here.
01:30
This is not right.
01:32
Let's go ahead and divide by seven.
01:35
And so this part here that i said in green earlier is actually, du over 7...