Closed form of $\int_0^x \frac{d y}{\sqrt{y^2+n^2}}\ln\frac{\sqrt{y^2+n^2}+y}{\sqrt{y^2+n^2}-y}$

Question

Let $n\in\mathbb{N}$. Consider the function $I_n:\mathbb{R}_+ \to \mathbb{R}$ defined as the definite integral $I_n(x)=\int_0^x \frac{1}{\sqrt{y^2+n^2}}\ln\frac{\sqrt{y^2+n^2}+y}{\sqrt{y^2+n^2}-y}dy$. How could $I_n(x)$ be expressed in a closed form? Thank you for your trouble!

Answer 1

The integral can be written as $$I_n=\int_0^x \frac{\log \left(\sqrt{n^2+y^2}+y\right)-\log \left(\sqrt{n^2+y^2}-y\right)}{\sqrt{n^2+y^2}}\, dy$$

Substitute $\sqrt{n^2+y^2}+y=t$ and as $y>0$ we get $y=\frac{t^2-n^2}{2 t}$

Differentiating $dy=\dfrac{n^2+t^2}{2 t^2} \,dt$

The interval of integration becomes $\left(n;\;x+\sqrt{n^2+x^2}\right)$

The integral becomes $$I_n=-2 \log n\int_n^{x+\sqrt{n^2+x^2}} \frac{dt}{t} =2 \log n \left(\log n-\log \left(x+\sqrt{n^2+x^2}\right)\right)$$

Hope this helps