Trouble integrating xe^(xy) for dy and dx

[Aegis]

I'm in a calc-based stats course, but I haven't had calc for quite awhile and I can't remember how to go about problems below. The question wants me to find the marginal probability distributions for f(x,y), which translates to the these two integrals:

I know

For the top one, I treat x as a constant, and it should work out to the same thing, correct?

For the bottom one, I've no idea where to begin. It seems like I'd want to break it up into (x)(e^(-xy-x)) but I'm unsure. Phrases like "product rule" "chain rule" "substitution" are floating around in my head, but I can't remember any of them, and so I don't even know where to look to educate myself, short of starting in Chapter one of my old calc book.

 

[galactus]

I think you're trying to recollect 'Integration by Parts'.

For the first one. The toughest one, wrt x.

\(\displaystyle \L\\\int{xe^{-x(1+y)}}dx\)

Let \(\displaystyle \L\\u=x; \;\ dv=e^{-x(1+y)}dx; \;\ du=dx; \;\ v=\frac{-e^{x(-y-1)}}{y+1}\)


\(\displaystyle \L\\udv-\int{vdu}\)

\(\displaystyle \L\\\frac{-xe^{x(-y-1)}}{y+1}+\int\frac{e^{x(-y-1)}}{y+1}dx\)

\(\displaystyle \L\\\frac{-xe^{x(-y-1)}}{y+1}+\frac{e^{x(-y-1)}}{(y+1)^{2}}\)

Factor:

\(\displaystyle \H\\\left(\frac{-x}{y+1}-\frac{1}{(y+1)^{2}}\right)e^{x(-y-1)}\)