Evaluate the integral: I = integral (sqrt x)/((1+ sqrt

Evaluate the integral: $$ I = \int \frac{\sqrt x}{\left(1+\sqrt x\right)^2}\,dx. $$

Explanation of Method

The substitution $u=\sqrt x$ turns the integrand into $2u^2/(1+u)^2$, which polynomial division reduces to $2\left[1-\tfrac2{1+u}+\tfrac1{(1+u)^2}\right]$, a sum of elementary pieces.

Substitute $u=\sqrt x$.
Divide the rational function.
Integrate term by term.
Back-substitute.

Step 1 · Substitute

Let $u=\sqrt x$, so $x=u^2$, $dx=2u\,du$:

$$ I = \int\frac{u}{(1+u)^2}\cdot2u\,du = 2\int\frac{u^2}{(1+u)^2}\,du. $$

Step 2 · Divide

Writing $u^2=(1+u)^2-2(1+u)+1$:

$$ \frac{u^2}{(1+u)^2} = 1 - \frac2{1+u} + \frac1{(1+u)^2}. $$

Step 3 · Integrate

$$ I = 2\left[u - 2\ln|1+u| - \frac1{1+u}\right] + C = 2u - 4\ln|1+u| - \frac2{1+u} + C. $$

Step 4 · Back-substitute

With $u=\sqrt x$:

Summary Table

Piece	Antiderivative
$1$	$u$
$-2/(1+u)$	$-2\ln\|1+u\|$
$1/(1+u)^2$	$-1/(1+u)$

Additional Notes

The domain is $x>0$.
The result was verified by differentiation at $x=4$: the derivative matches the integrand $\sqrt x/(1+\sqrt x)^2$ exactly.

Result

The solution to the integral is:

$$ I = 2\sqrt x - 4\ln\left(1+\sqrt x\right) - \frac2{1+\sqrt x} + C. $$