# Integration of Partial Fraction Expansion

Hi This is my first time posting a question on this website. Thank you advance for helping me out here.

My question is

Suppose the density of $X$ is $$f(x) = \frac{Kx^2}{(1 + x)^5}$$ when $x > 0$. Find the constant $K$ and the density of $Y = \frac{1}{(1 + X)}$.

++one more thing since this is pdf. This equals to 1.

I tried partial fraction expansion to do it but it was long and didn't lead to answer. For the density part I finished upto $\frac{-Kx^2}{(x+1)^3}$. However, I do not know K and could not finish it.

Thank you again

• $f(x) = Kx^2=(1 + x)^5$?
– msm
Oct 1 '16 at 7:40
• sorry its Kx^2/(1+x)^5 Oct 1 '16 at 7:41

Hint: Use $u$-sub first to get \begin{align} \int \frac{Kx^2}{(1+x)^5}\ dx = \int \frac{K(u-1)^2}{u^5}\ du. \end{align}