Find $\mathbb E[(X+1)^2]$

Question

The continuous random variable $X$ has the following probability density function (pdf),

$$f(x) =\frac{3}{(1 + x)^3} ; 0 ≤ x ≤\sqrt3-1$$

I have to find $\mathbb E[(X+1)^2]$

$$\mathbb E[(X+1)^2]=\int_0^{\sqrt3-1}(x+1)^2 \frac{3}{(1 + x)^3}dx=3\int_0^{\sqrt3-1}\frac{1}{(1 + x)}dx$$

Then i couldn't solve it.

Answer 1

Note that $\frac{d(\ln x)}{dx}=\frac{1}{x}$. So then you get:

$$ 3\int_0^{\sqrt{3}-1}\frac{1}{1+x}dx=3\left[\ln (x+1)\right]_0^{\sqrt{3}-1}=3\ln(3^{1/2})=\frac{3\ln(3)}{2}. $$