What is the indefinite integral of {[square root(x+1)] +1}^(-1)?

Let f(x) =
1/[1+sqrt(x+1)]

Int f(x) dx =Int
dx/[1+sqrt(x+1)]

We'll replace sqrt(x + 1) by
t.

We'll raise to square both
sides:

x + 1 = t^2

x = t^2 -
1

We'll differentiate both
sides:

dx = 2tdt

We'll
re-write the integral in t:

Int 2tdt/(1+t) = 2Int
tdt/(1+t)

We'll add and subtract 1 to numerator of the
ratio and we'll split the fraction in 2 fractions, using the property of integral to be
additive:

2Int (t + 1 - 1)dt/(1+t) = 2Int (t+1)dt/(1+t) -
2Int dt/(t+1)

We'll simplify and we'll
get:

Int f(x) dx = 2Int dt - 2ln |t +
1|

We'll change the variable
t:

Int f(x) dx = 2*sqrt(x + 1) - 2ln [sqrt(x + 1) + 1] +
C

Int f(x) dx = 2*{sqrt(x + 1) - ln [sqrt(x +
1) + 1]}+ C