Moments of Transmuted Lindley Distribution I've been having difficulty in obtaining the exact solution given by the author. I need help as my solution doesn't factorize completely to the same term given by the author.
Obtaining the kth moments entails computing the integral given in the picture. I carried out the integration but I did not get the exact term the author got. I was of the opinion that I may be wrong in my factorization, hence I seek other ideas to obtain the exact solution the author got. Otherwise, I want to be justified that the author is probably wrong.


 A: Focus on the form of the integrand that matches what you can compute.  Specifically, the expectation can be written
$$\eqalign{
\mathbb{E}(X^k) &= A\int_0^\infty x^k(1+x)e^{-\theta x}\left(B + (C+Dx)e^{-\theta x}\right)\mathrm{d}x \\
&=A\int_0^\infty\left((Bx^k + Bx^{k+1})e^{-\theta x} + (Cx^k + (C+D)x^{k+1} + Dx^{k+2})e^{-2\theta x}\right)\mathrm{d}x
}$$
with $A = \theta^2/(1+\theta)$, $B=1-\lambda$, $C=2\lambda$, and $D=2\lambda\theta/(1+\theta)$.
This splits into a sum of five gamma integrals, so we can read off the answer directly: it is
$$A\left(Bk!\theta^{-k-1} + B(k+1)!\theta^{-k-2} + Ck!(2\theta)^{-k-1} + (C+D)(k+1)!(2\theta)^{-k-2} + D(k+2)!(2\theta)^{-k-3}\right).$$
Factor out $\theta^{-k-2}k!$ to obtain
$$\frac{Ak!}{\theta^{k+2}}\left(B(\theta+k+1) + C2^{-k-1}\theta + (C+D)(k+1)2^{-k-2} + D(k+2)(k+1)2^{-k-3}\theta^{-1}\right).$$
The term involving $B$ equals
$$\frac{k!}{\theta^k(1+\theta)}(1-\lambda)(\theta+k+1),$$
recognizable in the first part of the answer.  The other terms equal
$$\frac{k!\lambda}{\theta^k(1+\theta)^2 2^{k+2}}\left(k^2 + (4\theta+5)k + 4(1+\theta)^2\right).$$
That's not remotely like what the author obtained.  Although I, too, may have made some algebraic mistake, it is obvious that some multiple of $k^2$ must be involved, because the original integral includes a multiple of $x^{k+2}e^{-2\theta x}$, which will introduce $(k+2)!=(k^2+3k+2)k!$ upon integration: that $k^2$ cannot be canceled off by any of the other terms.
