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I can't figure out how to solve for ∫ sqrt(2x^2-4) / (7x). Can someone please help me?

 Feb 26, 2019
 #1
avatar+99272 
0

2x^2-4>0

x^2-2>0

x^2>2

x>sqrt2    or   x<-sqrt2

 

I have not worked it out but I am thinking that you could let    \(\theta=sec^{-1}\frac{x}{\sqrt2}\)

 

that should help.    wink

 Feb 26, 2019
edited by Melody  Feb 26, 2019
 #2
avatar+27531 
+3

"I can't figure out how to solve for ∫ sqrt(2x^2-4) / (7x). Can someone please help me?"

 

Like so:

 

 

(As it is an indefinite integral you should add an arbitrary constant of course.)

.

 Feb 26, 2019
edited by Alan  Feb 26, 2019
 #3
avatar+98091 
0

Very nice, Alan!!!!

 

 

cool cool cool

CPhill  Feb 26, 2019

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