Let $$f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$$ $$(x \geq 0)$$. Then $$f(3) - f(1)$$ is equal to:

We have to evaluate the difference $$f(3)-f(1)$$ where$$\displaystyle f(x)=\int\frac{\sqrt{x}}{(1+x)^2}\,dx,\qquad x\ge 0.$$

To simplify the integrand we put $$t=\sqrt{x}\;.$$ Then $$x=t^2\quad\text{and}\quad dx=2t\,dt.$$ Substituting these in the integral, we obtain

$$\int\frac{\sqrt{x}}{(1+x)^2}\,dx \;=\;\int\frac{t}{(1+t^2)^2}\,(2t\,dt) =\int\frac{2t^2}{(1+t^2)^2}\,dt.$$

Therefore

$$f(3)-f(1)=\int_{x=1}^{x=3}\frac{\sqrt{x}}{(1+x)^2}\,dx =\int_{t=1}^{t=\sqrt3}\frac{2t^2}{(1+t^2)^2}\,dt.$$

Next we rewrite the integrand algebraically. Notice that

$$2t^2=2\bigl(t^2+1-1\bigr)=2(t^2+1)-2,$$

so that

$$\frac{2t^2}{(1+t^2)^2}= \frac{2(t^2+1)}{(1+t^2)^2}-\frac{2}{(1+t^2)^2} =\frac{2}{1+t^2}-\frac{2}{(1+t^2)^2}.$$

Thus

$$\int_{1}^{\sqrt3}\frac{2t^2}{(1+t^2)^2}\,dt =\int_{1}^{\sqrt3}\frac{2}{1+t^2}\,dt-\int_{1}^{\sqrt3}\frac{2}{(1+t^2)^2}\,dt.$$

We evaluate the two integrals separately.

First integral: we recall the standard result $$\displaystyle\int\frac{1}{1+t^2}\,dt=\arctan t.$$ Hence

$$\int\frac{2}{1+t^2}\,dt=2\arctan t.$$

Second integral: we need $$\displaystyle\int\frac{1}{(1+t^2)^2}\,dt.$$ To obtain it, we use two known derivatives:

$$\frac{d}{dt}\bigl(\arctan t\bigr)=\frac{1}{1+t^2},$$ $$\frac{d}{dt}\!\left(\frac{t}{1+t^2}\right)=\frac{1-t^2}{(1+t^2)^2}.$$

Adding these derivatives and dividing by 2 gives

$$\frac{1}{2}\left[\frac{1-t^2}{(1+t^2)^2}+\frac{1}{1+t^2}\right] =\frac{1}{(1+t^2)^2}.$$

Integrating both sides yields the useful identity

$$\int\frac{1}{(1+t^2)^2}\,dt=\frac{1}{2}\left(\frac{t}{1+t^2}+\arctan t\right)+C.$$

Multiplying by 2, we get

$$\int\frac{2}{(1+t^2)^2}\,dt=\frac{t}{1+t^2}+\arctan t.$$

Putting the two evaluated parts together, the antiderivative of our integrand is

$$2\arctan t-\bigl(\,t/(1+t^2)+\arctan t\,\bigr) =\arctan t-\frac{t}{1+t^2}+C.$$

Remembering that $$t=\sqrt{x},$$ we may write

$$f(x)=\arctan(\sqrt{x})-\frac{\sqrt{x}}{1+x}+C.$$

Since the constant $$C$$ cancels in the required difference, we compute directly:

For $$x=3$$ we have $$\sqrt{3}$$ and $$\arctan(\sqrt3)=\dfrac{\pi}{3},$$ so

$$f(3)=\frac{\pi}{3}-\frac{\sqrt3}{1+3} =\frac{\pi}{3}-\frac{\sqrt3}{4}.$$

For $$x=1$$ we have $$\sqrt{1}=1$$ and $$\arctan(1)=\dfrac{\pi}{4},$$ so

$$f(1)=\frac{\pi}{4}-\frac{1}{1+1} =\frac{\pi}{4}-\frac12.$$

Subtracting,

$$f(3)-f(1)=\left(\frac{\pi}{3}-\frac{\sqrt3}{4}\right) -\left(\frac{\pi}{4}-\frac12\right) =\frac{\pi}{3}-\frac{\pi}{4}+\frac12-\frac{\sqrt3}{4}.$$

Now $$\displaystyle\frac{\pi}{3}-\frac{\pi}{4} =\frac{4\pi-3\pi}{12}=\frac{\pi}{12},$$ so finally

$$f(3)-f(1)=\frac{\pi}{12}+\frac12-\frac{\sqrt3}{4}.$$

Hence, the correct answer is Option D.