Given this $p.d.f$$\rightarrow \displaystyle\frac{1}{\lambda}\frac{\theta x^{\theta -1}}{(1+x^{\theta})^{1/\lambda +1}}$ ,$x \in (0,\infty)$
I should arrive at an expression of this form: $\displaystyle h(x)a(\theta)exp[{\sum_{i=1}^{p}c_{i}(\theta)r_{i}(x)}]$
Operating the p.d.f $\displaystyle{\rightarrow \frac{1}{\lambda}\exp[(\theta -1)\ln(\theta x)-(\frac{1}{\lambda +1}) \ln (1+x^{\theta})]}$
But I got this far and I can't continue factoring to get to the expression I want.
The p.d.f should be this instead $\displaystyle{\rightarrow \frac{1}{\lambda} \theta \exp[\ln(\theta x)-(\frac{1}{\lambda +1}) \ln (1+x^{\theta})]}$
[$ \ln (\theta x^{\theta-1} ) \neq (\theta-1) \ln x^{\theta x} $ ]
then $\ln(\theta x) = \ln(\theta) + \ln(x).$
and $\exp((\frac{1}{\lambda +1}) \ln (1+x^{\theta})) = \exp(1/(\lambda+1)) (1+x^\theta) = \exp(1/(\lambda+1)) +\exp(1/(\lambda+1)) x\exp \theta $, since $\exp(ln(x^\theta)) = e^\theta x$.
Combine that and I believe that should lead you to the answer. Also, be careful of $1/(\lambda+1)$ instead of $1/\lambda +1$. Your problem statement says one and the solutions says the other.