How do you integrate #int 1/sqrt(4x+8sqrtx-3) # using trigonometric substitution?

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Answer 1 · 2016-02-25T16:27:09

#t = sqrt(x)#

#dt = 1/(2sqrt(x))dx#

#dx = 2sqrt(x)dt#

#t^2=x#

#2intt/sqrt(4t^2+8t-3)dt#

canonic form of the denominator

#2intt/sqrt((2t+2)^2-7)dt#

factorize to have #1/k(t+1)^2-1# and do a substitution with #cos(u)# or #sin(u)# and you will get the result