I solving a problem of conditional distribution of $X$ given $Y=y$. I have solved prior steps and got to this integral solution:
$$F_{X\,\mid\,Y=y}(x) = \int_0^x \frac{z^2}{2y^3}e^{-z/y}dz $$
I am new to this and wondering if I can use list below: https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
or have to solve this directly such as $\frac {z^3}{6y^3}$ etc and plug in $x$ and $0$.
Thank you for your help.
On
\begin{align} & \int_0^x \frac{z^2}{2y^3}e^{-z/y}dz = \frac 1 2 \int_0^x \left( \frac z y \right)^2 e^{-z/y} \left( \frac{dz} y \right) = \frac 1 2 \int_0^{x/y} u^2 e^{-u} \, du \\[10pt] = {} & \frac 1 2 \int \Big( u^2\Big) \Big( e^{-u}\, du \Big) = \underbrace{\frac 1 2 \int w\,dv = \frac 1 2 \left( wv - \int v\, dw \right) }_\text{integration by parts} \\[10pt] = {} & \frac 1 2 \left( -u^2 e^{-u} - \int (-e^{-u}) (2u\,du) \right) = \left[ - \frac 1 2 u^2 e^{-u} \right]_{u:=0}^{u := x/y} + \int_0^{x/y} u e^{-u} \,du =\cdots \end{align}
You can handle the last integral as follows: \begin{align} \int u \Big( e^{-u}\,du\Big) = \underbrace{ \int u\,dv = uv - \int v\,du } _\text{integration by parts again} = \text{etc.} \end{align}
Hint. If one has to evaluate $$ \int_0^x z^2e^{-az}dz,\qquad a \ne 0, $$ one may perform an integration by parts in the following way, $$ \int_0^x z^2e^{-az}dz=\left[z^2 \cdot \frac{e^{-az}}{-a} \right]_0^x+\frac{1}{a}\int_0^x 2z\cdot e^{-az}dz $$ that is
$$ \int_0^x z^2e^{-az}dz=-\frac{x^2e^{-ax}}{a}+\frac{2}{a}\int_0^x z\cdot e^{-az}dz $$ then one more integration by parts allows one to conclude.
Can you finish it?