Activity part 1: int [ 2 / sqrt{x^2 + 6x + 5} ] dx = 2 int [ 1 / sqrt{(x+3)^2 - 4} ] dx = ?
- Thread starter armn91
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Dr.Peterson
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Why would you put two parts of the work on one problem in different threads???
There is an error in the last line of this first picture. What is [imath]\tan^2(\theta)-1[/imath] equal to? It is not [imath]\sec^2(\theta)[/imath]. You have the wrong substitution!
This is the second picture....
Ok, thank you for letting me know. I have reviewed it, so is this correct now?
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Yes, that is correct. Optional: you can simplify it further by folding the [imath]-2\ln 2[/imath], which is itself a constant, into the arbitrary constant [imath]C[/imath]
Because [imath]\displaystyle 2\ln\left|\frac{x+3+\sqrt{\left(x+3\right)^2-4}}{2}\right|=2\left[\ln\left|x+3+\sqrt{\left(x+3\right)^2-4}\right|-\ln2\right][/imath]
Because [imath]\displaystyle 2\ln\left|\frac{x+3+\sqrt{\left(x+3\right)^2-4}}{2}\right|=2\left[\ln\left|x+3+\sqrt{\left(x+3\right)^2-4}\right|-\ln2\right][/imath]