Is this really true :$\int_{1}^{n}\frac{dx}{\operatorname{\\erf(x)}}= n -\alpha$ with $\alpha$ is constant and $n >1$? This is $\int\frac{dx}{\operatorname{\\erf(x)}}$ a simple form integral which i can't to evaluate using variable change and integration by part to get it's form, but some computations in wolfram alpha coming up to me to know one side of it's behavior over the range $(1,n)$ with $n$ is a positive integer such that i have deduced this relation from some computation which i run :
$$\int_1^n \frac{dx}{\operatorname{\\erf(x)}}= n -\alpha$$  with         $\alpha \sim 0.944974758233 \cdots $ and $n >1$ , Now my question here is :
Is really what have got true ? and is this function can be expressed as elementary function ?
 A: If this was true then it would imply\begin{align*}\int_{n}^{n+1}\frac{dx}{\operatorname{\\erf}(x)}&= (n+1 -\alpha)-(n -\alpha)\\&=1\end{align*}
for all $n$, say greater than $1$.
Now this is wrong because $\left[\forall x\in[n,n+1],\frac{1}{\operatorname{\\erf}(x)}>1\right]\Longrightarrow \int_{n}^{n+1}\frac{dx}{\operatorname{\\erf}(x)}>1$
The reason yo got this result is that $\frac{1}{\operatorname{\\erf}(x)}$ is actually very close to $1$, so that a computer will not make the difference, if $n$ is large enough. Here large enough is not even that large.
A: Too long for a comment.
If you look here, you will find the nice approximation
$$\text{erf}(x)\approx\tanh \left(\sqrt{\pi }  \log (2)x\right)$$ which makes
$$\int \frac{dx}{\text{erf}(x)}\approx \frac{\log \left(\sinh \left(\sqrt{\pi } \log (2)x\right)\right)}{\sqrt{\pi } \log
   (2)}$$
$$I_n=\int_1^n \frac{dx}{\text{erf}(x)}\approx \frac{\log \left(\sinh \left(\sqrt{\pi }  \log (2)n\right)\right)}{\sqrt{\pi } \log
   (2)}-\frac{\log \left(\sinh \left(\sqrt{\pi } \log (2)\right)\right)}{\sqrt{\pi
   } \log (2)}$$ When $n$ becomes large $\sinh(n)\approx \frac {e^x}2$ and, for large $n$ 
$$I_n \approx n-\log (2)-\frac{\log \left(\sinh \left(\sqrt{\pi } \log
   (2)\right)\right)}{\sqrt{\pi } \log (2)}\approx n-1.05605$$
For $n=100$, numerical integration leads to $99.055$ while the above gives $98.944$
