00:01
Let's evaluate the integral of 1 over x plus a times x plus b.
00:07
Because a and b, or could be any real numbers, this problem will depend on whether or not a and b are equal.
00:15
Because if a equals b, for example, this means that we can write the integral as 1 over x plus a square, and that'll change how we go about the partial fraction to composition.
00:34
This will be something that the book would call case 2.
00:41
But really for us in this case, this basically is already in the form of a partial fraction to composition.
00:48
We just actually go ahead and integrate this with the u sub, use the power rule.
00:52
So this is why the case, it matters whether or not a equals b.
00:58
And we'll see in case 2 when a does not equal b, we'll have two distinct linear factors in the denominator.
01:04
So that's a different looking partial fraction to composition than what we have over here that we're just circling right now so first we're in this case a equals b we rewrite the integral and then you may be able into integrated how it is in one step but if this plus a is bothering you just go ahead and do a use of then you get d u equals the x so we could write this integral is one over u squared d u and then you could use the power rule to get negative 1 over u plus c and then go ahead replace u with x plus a.
01:57
So in case 1 if a equals b this should be the integral.
02:05
Now we have to also look at the case in which they're not equal so let's do this next.
02:11
So that's case 1 all over here boxed off in green.
02:17
Now for case 2, a and b are different, distinct, not equal.
02:26
So now we can go ahead and take the original fraction 1 over x plus a.
02:37
This form right here as it's in with a and b distinct, this is not a partial fraction to composition...