Estimate the area restricted by $f(x) = \log(x+1)/x \, , f(x-1), \, y=0$, and $y=a$. I need to estimate the area between the functions 
$f(x) = \log(x+1)/x \, , f(x-1), \, y=0$, and $y=a$. where $a>1$. 
Now I have tried quite a few ways to do this, but nothing comes to mind.
I tried writing out the taylor series, I tried changing this around. 
Nothing really gave a decent approximation. 
A decent in my mind would be anything greater than one decimal. Eg an error less than $E<10^2$.
Cheers =)
EDIT: After making a nice drawing it seems that if $a>>0$ then $A \approx a + \frac{1}{12}\pi^2$

 A: Short answer:
Any horizontal line $y=c$ with  $0< c\leq a$ intersects the  set $B$ between the two curves in a segment of length $1$. Therefore by Cavalieri's principle the area $\mu(B)$ is given by $\mu(B)=a$.
More info:
Cavalieri's principle (maybe known to Archimedes) is a $17^{\rm th}$ century version of the Fubini theorem. It says the following: When two three-dimensional bodies are cut by every horizontal plane in two equal areas then the two bodies have the same volume. (You need this principle to prove that the volume of any pyamid is given by $V={1\over3}G\cdot h$, where $G$ is the area of its  base and $h$ denotes its height.)
For a formal proof of $\mu(B)=a$ in your example let $$g:\quad y\mapsto x=g(y)\qquad(0<y< \infty)$$ 
be the inverse of the function $$f:\quad x\mapsto {\log(1+x)\over x}\qquad(-1<x<\infty)\ .$$
(This means: Look at your figure considering $y$ as independent vaiable.)
Then the domain $B$ in question is defined by
$$B:=\{(x,y)\ |\ 0<y\leq a,\ g(y)\leq x\leq g(y)+1\}\ .$$
It follows that the area  $\mu(B)$ can be computed as
$$\mu(B)=\int_B{\rm d}(x,y)=\lim_{\epsilon\to 0+}\int_\epsilon^a\bigl(\int_{g(y)}^{g(y)+1} \ dx\bigr)\ dy=\lim_{\epsilon\to 0+}\int_\epsilon^a 1\ dy=\lim_{\epsilon\to 0+}(a-\epsilon)=a\ ,$$
where the inner integration was with respect to $x$.
A: You can also use the Cavalieri integral directly. Assume that $g(y)$ is defined as in the above answer (inverse of $f(x)$). Then the cavalieri integral of the region is equal to $\int_{g(y)}^{g(y)+1}a dx=\lim_{c\rightarrow\infty}\int_{-1}^{0}a\cdot h'(x)dx=a\lim_{c\rightarrow\infty}\int_{-1}^0 1dx=a$. Where $h(x)=x-g(a)+c$ is the inverse transformation function and $c$ is the $x$-intercept of $g(y)$. Alternatively let $p(x) = x+g(a)-c$ be the transformation function then $\int_{g(y)}^{g(y)+1}a dx =\int_{g(y)}^{g(y)+1}a \cdot 1(x) dx =\lim_{c\rightarrow\infty}\int_c^{c+1}a.1(p(x))dx=\lim_{c\rightarrow\infty}\int_c^{c+1}adx=a$. See Cavalieri integration, QM 35(3) 2012 pp. 265-296 for more details.
