Evaluate: $I(\alpha)=\int_{0}^{\infty} \frac{\arctan\left ( \alpha (x-\operatorname{arsinh} x) \right ) }{x\sqrt{1+x^2} }\text{d}x$ I am interested this type integrals. Let
$$
I(\alpha):=\int_{0}^{\infty} \frac{\arctan\left ( \alpha 
(x-\operatorname{arsinh} x) \right ) }{x\sqrt{1+x^2} }\text{d}x.
$$
For example, some simple calculations showed that
$$
I\left ( \frac{1}{\pi}  \right )=-\pi\ln\left ( \frac{\sqrt{\pi} }{2}  \right ).
$$
My Question is: 
Are there any extended results about $I(\alpha)$? Any suggestion would be appreciated.
 A: I am very close to solution, but not exactly. Let,
$$I(t)=\int_{0}^{\infty}\frac{\arctan(\alpha(xt-\operatorname{arsinh} x))}{x\sqrt{x^2+1}}$$
Do substitution,
$$u=\operatorname{arsinh}x\implies du=\frac{dx}{\sqrt{1+x^2}}dx$$
$$I(t)=\int_{0}^{\infty}\frac{\arctan(\alpha(t\sinh u-u))}{\sinh u}du$$
Now differentiate on both sides wrt $t$
$$I'(t)=\int_{0}^{\infty}\frac{\operatorname{csch}u}{1+(\alpha)^{2}(t\sinh u-u)^{2}}\alpha \sinh u du$$
Notice that $\sinh u$ and $\operatorname{csch} u$ cancels off each other.
$$I'(t)=\alpha \int_{0}^{\infty}\frac{1}{1+(\alpha)^{2}(t\sinh u-u)^{2}}du$$
Now I am stuck here. The integral is simplified to some extent. But I have no clue what to do next. I tried to make use of $\frac{1}{1-x}=\sum_{k\ge 0}x^{k}, |x|<1$.
But using that brings us to something inside the integral which doesn't have an anti-derivative. And also if we can do the substitution $q=t\sinh u-u$ then I think something can be done. But, it's derivative is $t\cosh u-1$ which I am unable to express in terms of $q$.
