Given the density function of a certain population+
$$f_\theta(x)=\frac{x+1}{\theta(\theta+1)}e^{-\frac{x}{\theta}} \qquad \text{for } x>0, \ \theta>0$$
Since $f_\theta(x)$ is part of the exponential k-parametric family, from its factorization one immediately knows that $\bar X$ is a minimal sufficient statistic. Moreover, it is complete; this is a fact you may find in several statistical inference books.
However, I cannot use this result freely in my statistics course, and I may prove by definition that $\bar X$ is complete. Here is my attempt:
If one wants to prove that if $E[f(\bar X)]=0$, then one must have $f=0$ almost everywhere, first one has to explicitly give the density of $\bar X$ to compute the expectancy of $f(\bar X)$. However, it is not trivial what this density may be, as $f_\theta(x)$ does not correspond to the density of any of the well-known continuous distributions for which $\bar X$ is easily obtainable.
So, I may define the following change of variables:
$$(Y_1,Y_2,\ldots,Y_{n-1},Y_n)=(X_1,X_2,\ldots,X_{n-1},\bar X)$$
Which has as inverse function
$$(X_1,X_2,\ldots,X_n)=(Y_1,Y_2,\ldots,nY_n-(Y_1+Y_2+\cdots+Y_{n-1}))$$
And therefore, its jacobian matrix is
$$J=\begin{bmatrix} 1 & 0 & 0 & \dots & -1 \\ 0 & 1 & 0 & \dots & -1 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & n \end{bmatrix}$$
Which gives $|\det(J)|=n$. Therefore, one has
$$f_{(Y_1,\ldots,Y_n)}(y_1,\ldots,y_n)=nf_{(X_1,\ldots,X_n)}(y_1,\ldots,y_{n-1},ny_n-(y_1+\cdots+y_{n-1}))$$
where the latter equality is nothing but the joint sample density $f_\theta(x_1,\ldots,x_n)$ at the point $(y_1,\ldots,y_{n-1},ny_n-(y_1+\cdots+y_{n-1}))$
Subtituding one has
\begin{align} & f_{(Y_1,\ldots,Y_n)}(y_1,\ldots,y_n) \\ = {} & n\frac{(y_1+1)(y_2+1)\cdots(ny_n-(y_1+y_2+\cdots+y_{n-1}))}{\theta^n(\theta+1)^n}e^\frac{-(y_1+\cdots+y_{n-1}+ny_n-y_1-\cdots-y_{n-1})}{\theta} \\ = {} & n\frac{(y_1+1)(y_2+1)\cdots(ny_n-(y_1+y_2+\cdots+y_{n-1}))}{\theta^n(\theta+1)^n} e^{-\frac{ny_n}{\theta}} \end{align}
To finally obtain $f_{\bar X}(\bar X)$ one must calculate
$$f_{\bar X}(\bar X)=\int_{C_{Y_1}}\int_{C_{Y_2}}\cdots\int_{C_{Y_{n-1}}} f_{(Y_1,\ldots,Y_n)}(y_1,\ldots,y_n) \ dy_1 \, dy_2 \ldots dy_{n-1}$$
Which can be rewritten as
$$\frac{n}{\theta^n(\theta^n+1)}e^{-n\frac{y_n}{\theta}} \int_0^\infty \int_0^\infty\cdots\int_0^\infty(y_1+1)(y_2+1)\cdots(ny_n-y_1-\cdots-y_n-1) \ dy_1 \, dy_2\cdots dy_n$$
However, I don't know how to compute that integral, nor if my previous calculations are correct.
I think this density can also be found via obtaining the characteristic function, but I am not sure if that method is any better than finding the marginal distribution as I have just tried.
After all that one should still compute $\operatorname E[f(\bar X)]$; I hope its an easier task.
Thanks in advance for your collaboration.