I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:
$f(x) = c(x^2+1)\space\space\space if\space X \in [0,1];\space$ otherwise $0$
My attempt: $P(0\leq X \leq 1) = \int_0^1c*(x^2+1) = 1 \Rightarrow cx+\frac {cx^3}3 = 1 \Rightarrow c = 3/4$
The problem here is that when we put $x=1$ we get $f(1) = \frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.
The CDF cannot exceed $1$, but the PDF can. Your CDF is $\frac{3}{4}x+\frac{1}{4}x^3$ on $[0,\,1]$, just as you calculated.