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

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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

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

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

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

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

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

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

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