Answer
$C=2$ and $P(0 \leq X \leq 1)=\dfrac{3}{4}$
Work Step by Step
We are given that
$p(x)=\dfrac{C}{(x+1)^3}$
$\int_0^{\infty} \dfrac{C}{(x+1)^3} dx =\lim\limits_{a \to \infty}\int_0^{a} \dfrac{C}{(x+1)^3} dx\\=-\dfrac{C}{2}\lim\limits_{a \to \infty} \dfrac{1}{(a+1)^2} -1 \\=\dfrac{-C}{2}\lim\limits_{a \to \infty} \dfrac{-a^2-2a}{(a+1)^2}\\=\dfrac{C}{2}$
The condition will be satisfied when
$\int_0^{\infty} \dfrac{C}{(x+1)^3} dx=\dfrac{C}{2}=1$ or, $C=2$
We will find
$P(0 \leq X \leq 1)=\int_0^1 \dfrac{2}{(x+1)^3}dx=-\dfrac{1}{(x+1)^2}|_0^1 \\=\dfrac{3}{4}$
