How to find the antiderivative of the function f(x)=square root x/(1+square root x)?

We'll find the antiderivative of the given function
evaluating the indefinite integral of this function.

Int
f(x)dx = Int sqrtx dx/(1+sqrtx)

We'll solve the integral
using substitution technique. We'll substitute 1 + sqrt x =
t.

We'll differentiate both
sides:

dx/2sqrt x = dt

dx =
(2sqrt x)dt

But sqrt x = t -
1

dx = 2(t - 1)dt

We'll
re-write the integral;

Int f(x)dx = Int 2(t -
1)^2dt/t

We'll expand the
square;

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

We'll apply the property of integrals to be
additive:

2 Int (t^2 - 2t + 1)/t = 2Int t^2dt/t - 4Int
tdt/t + 2Int dt/t

2 Int (t^2 - 2t + 1)/t = 2Int tdt - 4 Int
dt + 2Int dt/t

2 Int (t^2 - 2t + 1)/t = 2*t^2/2 - 4t +
2ln|t| + C

We'll simplify and we'll
get:

2 Int (t^2 - 2t + 1)/t = t^2 - 4t + 2ln|t| +
C

Int f(x)dx = (1 + sqrt x)^2 - 4(1 + sqrt x)
+ ln (1 + sqrt x)^2 + C