Continuous random variables $X$ and $Y$ have joint density function:
$$ f_X,_Y(x,y) = \begin{cases} \frac{6x+3y^2}{4}, & \text{if 0 $\le$ x $\le$ 1 and 0 $\le$ y $\le$ 1 } \\[2ex] 0, & \text{otherwise} \end{cases} $$
$(a)$ Find a formula for the marginal density function $f_X(x)$. $\\$
What I have tried is the following to "cancel out the y":
$$ \int_{y=0}^{y=1} \frac{6x+3y^2}4 dy $$
The trouble I am having with this integral is how do I take out the x part to the outside of the integral?
I thought about $ \frac{6x+3y^2}{4} = \frac{3(2x+y^2)}{4} = \frac{3}{4}{(2x+y^2)}{} $ but then where do I go from here?
Just integrate $\frac{3(2x+y^2)}{4}$ wrt $y$ as $y$ goes from 0 to 1. You'll get the answer $\frac{6x+1}{4}$, which is $f_X(x)$.