Reason behind a property in conditional expectation

Question

If $E[X|Y=y]=\int_{0}^{\infty}xf_{X|Y}(x|y)dx=\int_{0}^{\infty}\frac{y+1}{\mu+1}f_{X|Y}(x|y+1)dx$, then why is $E[X|Y=y]=\frac{y+1}{\mu+1}$? Shouldn't it be $\frac{y}{\mu+1}$ since we are looking for $E[X|Y=y]$ and not $E[X|Y=y+1]$? I came across this in the answer key of part ii, b, problem 2 of this MIT OCW problem set.

Answer 1

They have shown that $xf_{X|Y}(x|y)=\frac{y+1}{\mu+1}f_{X|Y}(x|y+1)$, thus $$ E[X|Y=y]=\int xf_{X|Y}(x|y)dx=\int \frac{y+1}{\mu+1}f_{X|Y}(x|y+1)dx= \frac{y+1}{\mu+1}\int f_{X|Y}(x|y+1)dx=\frac{y+1}{\mu+1}\times1. $$ As $\frac{y+1}{\mu+1}$ does not depend on $x$, thus you just take an integration over a density function - that equals $1$.