So I more or less know how to do it, just don't know if what I did is right
Let $X = (X_1,...,X_n)$ be a random obeservation with density that follows:
$ \theta(1+x)^{(-1-\theta )} $
show that this distribution belongs to the exponential family.
What I got so far:
$$\Pi\;\theta(1+X_i)^{(-1-\theta )}$$
$$\theta^n\Pi(1+X_i)^{(-1-\theta)}$$
$$exp\{log \theta - (1+\theta)log (\Pi (1+X_i))\} $$
so with this I have it: $$h(x) = 1\;\; n(\theta)=(1+\theta)\;\; A(\theta) = nlog \theta \;\;T(x) = \Pi(1+X_i)$$
And for the natural exponential form;
with $\theta = e^{\mu}$
$$exp\{n\mu-(1+e^{\mu})log(\Pi(1+X_i)\}$$
Is this right?
A pdf/pmf belongs to exponential family if you can write it in the form $$c(\theta)h(x)\exp(\sum_{i=1}^{k}w_i(\theta)t_i(x))$$ So you have the pdf $$\theta(1+x)^{-(1+\theta)}$$ which can be written as $$\theta\exp(-(1+\theta)\log (1+x))$$ where $c(\theta)\equiv\theta$, $h(x)\equiv1$, $k=1$, $w_1(\theta)\equiv -(1+\theta)$ and $t_1(x)=\log (1+x)$. Hence it belongs to exponential family.