Integral of a product of functions divided by the integral of one of the two functions Ratio between two integrals: $$\frac {\int f(x)g(x)} { \int f(x)}.$$ 
Does exist a rule or do you know a way to solve it?
$f(x)= (1+x)^n  e^{-ax}$
$g(x)= \ln(1+x)$
So: the numerator is the integral of $f(x)g(x)$ and the denominator is only the integral of $f(x)$.
And the question is if there exists a solution of this ratio..

My complete problem is: $$\frac {\int f(x)g(x)} {\left( \int f(x)\right)^2}.$$  
 A: The first integral
$$\int{\ln(1+x)\left(1+x\right)^n e^{-ax}dx} $$
can  be solved in the following way:
$$\int{\ln(1+x)\left(1+x\right)^n e^{-ax}dx}=\left\{u=\ln(1+x);dv=(1+x)^{n}e^{-ax}dx\right\}=\ln(1+x)A(x)-\int\frac{1}{1+x}A(x)dx $$
Here
$$A(x)=\int\left(1+x\right)^n e^{-ax}dx $$
$A(x)$ can be found by n times integrating by parts
$$A(x)=\int\left(1+x\right)^n e^{-ax}dx =\frac{-1}{a}(1+x)^n e^{-ax}-\int n(1+x)^{n-1}e^{-ax}dx$$
The final answer is:
$$A(x)=\sum\limits_{k=0}^{n}\frac{(-1)^{k+1}}{a^{k+1}}\frac{n!}{(n-k)!}(1+x)^{n-k}e^{-ax} $$
Moreover,$$\int f(x)dx=A(x)$$
To find 
$$\int \frac{A(x)}{1+x}dx $$
you can repeat previous integration all times except only
$$ \int \frac{e^{-ax}}{1+x}dx$$
With the help of substitution it can be changed to something like
$$\int\frac{e^{-bx}}{x}dx .$$
But it is not evaluated in elementary functions.
A: In the case of definite integrals, we have the following theorem, called the weighted mean value theorem: If $f$ and $g$ are continuous on a closed interval $[a,b]$, and if $g$ never changes sign in $[a,b]$, then we have 
$$ \int_{a}^{b} f(x) \cdot g(x) \ dx = f(c) \cdot \int_{a}^{b} g(x) \ dx $$ for some $c$ in $[a,b]$. For reference, see Calculus vol. 1 by Tom M. Apostol, Theorem 3.16. Hope this piece of information is of help. 
A: I can not calculate the Integral
$$
\int f(x) g(x) \, dx
$$
But,
$$
\int f(x) \, dx=-e^{\alpha } (x+1)^{n+1} (\alpha  (x+1))^{-n-1} \Gamma (n+1,(x+1) \alpha )
$$
where $\Gamma(a,z)$ is incomplete gamma function.
This may be helpful for you.
