Q:

Help finding k. Issue with integration

Let the continuous random variable $X$ have a probability density function $f(x)$ such that
$$f(x) = k(1+x)^{-3}, x>0$$
$=0$ elsewhere
Find k
This is what I tried:
$\int_0^\infty k(1+x)^{-3}dx = 1$
since it's a probability density function
$k\int_0^\infty(1+x)^{-3}dx = 1$
$\int_0^\infty(1+x)^{-3}dx = \frac{1}{k}$
Let $u = 1+x$
$du = dx$
$\int_0^{\infty}u^{-3}du = {\frac{1}{k}}$
$[-\frac{u^{-2}}{2}]_0^{\infty} = \frac{1}{k}$
$[-\frac{(1+x)^{-2}}{2}]_0^{\infty} = \frac{1}{k}$
Now I'm stuck. Was I supposed to do something related to L'Hopitals Rule?

A:

You're doing it very well. Continuing your work, we will obtain
\begin{align}
-\frac12\lim_{a\to\infty}\left[\frac1{(1+x)^2}\right]_{x=0}^a&=\frac1k\\
\frac12&=\frac1k\\
k&=2
\end{align}
since
$$
\lim_{a\to\infty}\frac1{(1+a)^2}=0.
$$

