How do you find the indefinite integral of $int 1/sqrt(x+1)$ ?

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Answer 1 · 2017-01-12T00:44:20

Let $u = x + 1$ . Then $du = dx$ .

$=int 1/sqrt(u)du$

$= int u^(-1/2) du$

$=2u^(1/2) + C$

$=2sqrt(x + 1) + C$

Hopefully this helps!

How do you find the indefinite integral of int 1/sqrt(x+1)int 1/sqrt(x+1)?