How do you integrate #int (x+1)/sqrt(x-1)# using substitution?

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Answer 1 · 2016-07-19T10:28:42

#= 2/3 ( x + 5) sqrt(x-1) + C#

#int (x+1)/sqrt(x-1) \ dx#

we can try simple linear sub #u = x-1, x = u+1, dx = du#

integral becomes
#int (u+2)/sqrt(u) \ du#

#= int \ sqrt u+2/sqrt(u) \ du# which is very doable, just the power rule

#= 2/3 u^(3/2) +4 sqrt(u) + C#

#= 2/3 (x-1)^(3/2) +4 sqrt(x-1) + C#

which we can tidy up a bit.....

#= (2/3 x- 2/3 +4) sqrt(x-1) + C#

#= (2/3 x + 10/3) sqrt(x-1) + C#

#= 2/3 ( x + 5) sqrt(x-1) + C#

that's the simplest way i reckon