How do you evaluate the integral #int sqrtx/(x-1)#?

Question

1340 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T21:24:26

#= 2sqrtx + ln( (sqrtx-1) /(sqrtx +1) )+ C#

#int sqrtx/(x-1) dx#

it looks a little easier with a simplification: #y = sqrt x, dy = 1/(2 sqrt x) dx implies dx = 2 y dy#

giving us

#2 int ( y^2)/(y^2-1) dy#

#=2 int ( y^2 - 1 + 1)/(y^2-1) dy#

#=2 int 1 + ( 1)/(y^2-1) dy =2 int 1 + ( 1)/((y-1)(y +1)) dy#

#=2 int 1 + 1/2( 1/(y-1) - 1/(y +1)) dy#

#= 2y + ln( (y-1) /(y +1) )+ C#

#= 2sqrtx + ln( (sqrtx-1) /(sqrtx +1) )+ C#