Value of improper Integral I need help in finding the value of the integral $$\displaystyle \int_0^\infty \left(\frac{x^2}{1+x}\right)^{n-1}e^{-tx}dx,$$ where $n$ is a positive integer and $t$ is a positive real number.
 A: The answer can be expressed with a confluent hypergeometric function \begin{equation}
\int_0^\infty \left(\frac{x^2}{1+x}\right)^{n-1}e^{-tx}\;dx, = \Gamma(2n-1)U(2n-1;n+1;t)
\end{equation}
for $\text{Re}(t)>0$ and $\text{Re}(n)>\frac{1}{2}$. 
One way to show this is a Mellin transform under the integral sign with respect to the parameter $t$ to remove the exponential 
$$
\int_0^\infty \mathcal{M}_t\left[\left(\frac{x^2}{1+x}\right)^{n-1}e^{-tx}\right]\;dx= \mathcal{M}_t[I_n(t)]
$$
$$
\int_0^\infty \left(\frac{x^2}{1+x}\right)^{n-1}\underbrace{\int_0^\infty \;t^{s-1} e^{-tx}dt}_{\Gamma(s)/x^s} \;dx= \mathcal{M}_t[I_n(t)]
$$
then $$
\Gamma(s)\int_0^\infty x^{-s}\left(\frac{x^2}{1+x}\right)^{n-1} \;dx= \mathcal{M}_t[I_n(t)]
$$
$$
\frac{\Gamma(2n-s-1)\Gamma(s)\Gamma(s-n)}{\Gamma(n)}= \mathcal{M}_t[I_n(t)]
$$
then perform an inverse Mellin transform to both sides 
$$
\frac{1}{2 \pi i}\int_{-i\infty}^{i\infty} t^{-s}\frac{\Gamma(2n-s-1)\Gamma(s)\Gamma(s-n)}{\Gamma(n)}\;ds=I_n(t)
$$
which when interpreted as a Barnes integral gives 
$$
I_n(t)=\Gamma(2n-1)U(2n-1;n+1;t)
$$
