00:01
Okay, we've been asked to use the u of square root of x minus 2, do our u substitution, and then see how we can integrate this.
00:12
So if u is what's given, we can go ahead and find du in terms of dx.
00:19
So i'm going to go ahead and get our formulas to the side here for when we're ready for our inverse trig formulas.
00:27
Okay, so our du is going to be one -half.
00:31
Of x minus 2 now it's to the negative one half power and then the derivative the inside is just d x so if we multiply both sides by that x minus 2 the half power that really is u and we also multiply the 2 up so we end up with a 2 u d u equaling d x now we also have that x plus 1 in the denominator that also has to be in terms of you.
01:09
So if i take my initial u statement, i square both sides and then i add three to both sides, now i see that x plus one can be replaced with a u squared plus three.
01:26
Okay, now that we have our u substitution, we see that we have the same degree in the numerator and the denominator.
01:32
So we're going to go ahead and we're going to do long division because when using our integration techniques, we need to have inside our integral...