Laplace Transform of $f(x)=\frac{\sqrt{x}}{1+x}$ What is the Laplace transform of this function? $$f(x)=\frac{\sqrt{x}}{1+x}$$
 A: We want  $\displaystyle F(t)=\int^{\infty}_0 \frac{\sqrt{x}}{1+x} e^{-t x} dx$
$F(t) e^{-t}=\int^{\infty}_0 \frac{\sqrt{x}}{1+x} e^{-t(x+1)} dx$
$\frac{d}{dt}(F(t) e^{-t})=-\int^{\infty}_0 \sqrt{x}e^{-t(x+1)} dx$
$e^t\frac{d}{dt}(F(t) e^{-t})=-\int^{\infty}_0 \sqrt{x}e^{-tx} dx=-t^{-1/2-1}\Gamma(1/2+1)=-\sqrt{\pi}/(2t^{\frac32})$
because you got a specific case of the integral defining the Gamma function.
After that you'll have to revert the operations :
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C+\int \frac{e^{-t}}{t\sqrt{t}} dt\right)$
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C-\frac{2e^{-t}}{\sqrt{t}}-\int \frac{2e^{-t}}{\sqrt{t}} dt\right)$
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C-\frac{2e^{-t}}{\sqrt{t}}-2\int e^{-u^2} du\right)$
where we recognize the integral expression of the Error function multiplied by $\sqrt{\pi}$.
I'll let you reverify all this and determine the constant $C$.
The answer should be $\sqrt{\frac{\pi}{t}} -\pi e^t \text{ erfc}(\sqrt{t})$  for $\Re(t)\gt0$ (with $\text{erfc}(x)=1-\text{erf}(x)$).
(short way : Wolfram Alpha).
A: $$ f(t)=\frac{\sqrt{t}}{1+t}  $$
$$ (1+t)f(t)=\sqrt{t}   $$
 $$ \downarrow {\mathcal{L}} $$ 
$$ F(s)-\frac{d}{ds}F(s)=\frac{\frac{\sqrt{\pi}}{2}}{s^{\frac{3}{2}}} $$
we have First Order Differential Equation:
 $$ \frac{d}{ds}F(s)-F(s)=-\frac{\frac{\sqrt{\pi}}{2}}{s^{\frac{3}{2}}}  $$
We know :
$$ \dfrac{dy}{dt}+ P(t) y = Q(t)  \space , \space y=e^{-\int P(t)dt} \left[\int Q(t) e^{\int P(t)dt} dt+c \right]  $$
So :
$$ F(s)=e^{-\int (-1)ds}\left[ \int \frac{\frac{-\sqrt{\pi}}{2}}{s^{\frac{3}{2}}}e^{\int(-1)ds}ds+C \right] $$
$$ F(s)=e^s \left[ (\frac{-\sqrt{\pi}}{2})  \int \frac{1}{s^{\frac{3}{2}}} e^{-s} ds +C\right] $$
$$ F(s)=e^s \left[ (\frac{-\sqrt{\pi}}{2}) \left( \frac{-2e^{-s}}{\sqrt{s}} -2\sqrt{\pi} .Erf (\sqrt{s}) \right) +C \right]$$
$$ F(s)=\sqrt{\frac{\pi}{s}}+\pi e^s.Erf(\sqrt{s})+Ce^s $$
