By Dawkins P.
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Extra info for Calculus II
In this case we do have an x in the numerator however the numerator still isn’t a multiple of the derivative of the denominator and so a simple Calculus I substitution won’t work. aspx Calculus II So, let’s again complete the square on the denominator and see what we get, x 2 + 10 x + 28 = x 2 + 10 x + 25 − 25 + 28 = ( x + 5 ) + 3 2 Upon completing the square the integral becomes, ⌠ 3x − 1 3x − 1 ⌠ dx = dx 2 2 ⌡ x + 10 x + 28 ⌡ ( x + 5) + 3 At this point we can use the same type of substitution that we did in the previous example.
Make sure that you can do those integrals. There is also another integral that often shows up in these kinds of problems so we may as well give the formula for it here since we are already on the subject. 1 1 x ⌠= dx tan −1 + c 2 2 ⌡ x +a a a It will be an example or two before we use this so don’t forget about it. Now, let’s work some more examples. Example 2 Evaluate the following integral. x2 + 4 ⌠ dx 3 ⌡ 3x + 4 x 2 − 4 x Solution We won’t be putting as much detail into this solution as we did in the previous example.
The first list is integration techniques that simply won’t work and the second list is techniques that look like they might work. After going through the strategy and the second list has only one entry then that is the technique to use. If, on the other hand, there are more than one possible technique to use we will then have to decide on which is liable to be the best for us to use. Unfortunately there is no way to teach which technique is the best as that usually depends upon the person and which technique they find to be the easiest.
Calculus II by Dawkins P.