Activity part 1: int [ 2 / sqrt{x^2 + 6x + 5} ] dx = 2 int [ 1 / sqrt{(x+3)^2 - 4} ] dx = ?

[armn91]

Hello, could you please let me know if what I am doing is correct ? There are two pictures, I will upload the second one in another thread. Thank you!!

 

[Dr.Peterson]

Hello, could you please let me know if what I am doing is correct ? There are two pictures, I will upload the second one in another thread. Thank you!!

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!

 

[armn91]

This is the second picture....

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!

Ok, thank you for letting me know. I have reviewed it, so is this correct now?

 

[limiTS]

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]

 

[armn91]

Oh, ok, thank you very much!