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

Question

1531 views around the world

You can reuse this answer
Creative Commons License

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

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