Integration by parts, three times I have a basic problem with integration by part, I know the answer to c is 1/8 per solution. But I can't carry out the steps, professor mentioned intergration by parts 3 times. Can you please show me how to go from the blue to (c=1/8)?
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 A: 
I hope this is clear. I skipped details of evaluating the limits as they approach infinity (this can be done on your part) but I have the 3 "integration by parts" steps that you were seeking.
A: Hopefully I am getting your question correctly. You are trying to calculate
$$I:=\int\mathrm e^{-y} y^3 \; \mathrm d y.$$
Set $u'= \mathrm e^{-y}$ and $v=y^3$. Integration by parts gives $\int u'v= uv - \int u v' $ that is 
$$ I=-\mathrm e^{-y} y^3 + 3\int  \mathrm e^{-y} y^2 \; \mathrm d y. $$
Now you have to integrate 
$$I_1:=\int  \mathrm e^{-y} y^2 \;\mathrm d y$$
by parts again which gives you
$$I_1 = -\mathrm e^{-y} 3y^2 + 2\int  \mathrm e^{-y} y \;\mathrm d y. $$
Set 
$$I_2 :=\int  \mathrm e^{-y} y \;\mathrm d y$$
and integrate again by parts. Hence you get
$$I_2= -\mathrm e^{-y} y+ \int \mathrm e^{-y} \; \mathrm d y = -\mathrm e^{-y} (y+1) .$$
Finally you get (collecting all the results)
$$I = -\mathrm e^{-y} \left(y^3 +2y^2 +6x +6 \right)$$
and 
$$\int_0^\infty \mathrm e^{-y} y^3 \; \mathrm d y=6.$$
Hint: In my opinion it is better to put 
$$ -e^{-y} (ay^3 +by^2 +cy +d)$$
as a possible primitive and solve for the coefficients $a,b,c,d$.
