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

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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**

To find the anti derivative of f(x) = x^(1/2)/(1+x^(1/2).

We put x^1/2 = t. Then by differentiating x^(1/2) = t, we get:

(1/2) dx /x^(1/2) = dt. So dx = 2t*dt.

Therefore f(x) dx = {x^(1/2)/(1+x^(1/2))}dx = Int(t/(1+t))2tdt = 2t^2/(t+1).

f(x) dx = Int 2t^2/(1+t) = 2Int{{(t+1)t - (t+1) +1}/(t+1) }dt.

f(x) dx = 2 {Int t dt -1Int dt + Int (1/1+t) dt}.

f(x) dx = 2 { t^2/2 - t + ln (1+t) }+ C.

f(x) dx = t^2 - 2t +2 ln (1+t^(1/2)) +C.

f(x) = (1+sqrtx)^2 - 2(1+sqrtx) +2ln (1+sqrtx) + C.

Int (x^(1/2)/ (1+x^(1/2)) = 1+sqrtx)^2 - 2(1+sqrtx) +2ln (1+sqrtx) + C.