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!
I'm unable to upload several images, so here are Flickr links to my work:
http://www.flickr.com/photos/99086960@N07/9317018784/
http://www.flickr.com/photos/99086960@N07/9317018894/
1 Answer
mathsworkmusic | (Level 2) 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.