When solving the following integral, Wolfram Alpha tells me this:integral (sqrt(1/2 (x+sqrt(x^2-3)))+sqrt(1/2 (x-sqrt(x^2-3)))) dx = 1/3 sqrt(2) (sqrt(x-sqrt(x^2-3)) (sqrt(x^2-3)+2...

When solving the following integral, Wolfram Alpha tells me this:

integral (sqrt(1/2 (x+sqrt(x^2-3)))+sqrt(1/2 (x-sqrt(x^2-3)))) dx = 1/3 sqrt(2) (sqrt(x-sqrt(x^2-3)) (sqrt(x^2-3)+2 x)-(sqrt(x^2-3)-2 x) sqrt(sqrt(x^2-3)+x))+constant

However, I don't understand how the √2 factors out of the expression or where the coefficient of 2 comes in for x in the parts (sqrt(x^2-3)+2 x) and (sqrt(x^2-3)-2 x). I've attached my work as an image. Could you please point out at what step in the process I've made a mistake? Or maybe it's that I'm missing a step at the end. I realize that the work goes into unnecessary expansions at some points. This is because I am trying to explain the process step by step to someone.
Thanks in advance!

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Expert Answers
mathsworkmusic eNotes educator| Certified Educator

You are forgetting that the subfunction `sqrt(x^2-3)` needs to be integrated also.

Realising that the derivative of

`1/3(x pm sqrt(x^2-3))^(3/2) = 1/2(x pm sqrt(x^2-3))^(1/2)(1 pm (2x)/sqrt(x^2-3)) `

` ` `= ((sqrt(x^2-3)pm 2x)/sqrt(x^2-3))1/2sqrt(x pm sqrt(x^2-3))`

gives what `u` and `v` would be when using the method of integrating by parts. ie use

`u = sqrt(x^2-3)/(sqrt(x^2-3) pm 2x)`

`v = 1/3(x pm sqrt(x^2-3))^(3/2)`

That explains where the `2x` comes in but I haven't gone through this fully to see if this is a viable method to use.