Find an integral $\DeclareMathOperator{\arcsinh}{arcsinh}
\DeclareMathOperator{\sgn}{sgn}
\newcommand{\dd}{\mathop{}\!\mathrm{d}}$
My integral follows:
$$\int\limits_0^n\left(\int\limits_0^n \dfrac{1}{\sqrt{x^2+y^2}+1}\dd x\right)\dd y.$$
I attempted the following:
$$(\text{integral above})=\int\limits_0^n\left(\int_0^1\dfrac{1}{|y|\sqrt{t^2+1}+1}y\dd t\right)\dd y,$$
by substituting $t=\frac{x}{y}$, whence $\dd x=y\dd t$. Let's have a look at that internal integral. The root suggests a hyperbolic substitution $t=\sinh z$, whence $\dd t=\cosh z\dd z$. That leads to:
$$\int\limits_0^{\arcsinh(1)}\dfrac{1}{|y|\cosh z+1}y\cosh z\dd z=\sgn(y)\left(\int\limits_0^1\dfrac{|y|\cosh(z)+1-1}{|y|\cosh(z)+1}\dd z\right)=$$
$$=\sgn(y)\arcsinh\Big(\frac{x}{y}\Big)\Big|_0^n-\sgn(y)\int\limits_0^{\arcsinh(1)}\dfrac{1}{|y|\cosh(z)+1}\dd t.$$
That is where I got stuck. The first term clearly evaluates to $\sgn(y)\arcsinh(\frac{n}{y})$, but then how do I integrate it in $\dd y$? And how do I integrate the second term? Is there an easier way to do this?
 A: This was once an edit to the question, but it should be an answer, with opportune completions.

Edit:
Taking the comments into account, here is another approach.
Using polar coordinates, the integral becomes:
  $$\int\limits_0^{\frac{\pi}{2}}\left(\int\limits_?^?\dfrac{r}{1+r}\mathrm{d}r\right)\mathrm{d}\theta.$$
  By symmetry, we turn this into twice the integral with theta stopping at $\frac{\pi}{4}$. With that, the inner limits are 0 and $\frac{n}{\cos(\theta)}$. In the other half, the cosine in the limit would be a sine. Thus the integral is:
  $$2\cdot\int\limits_0^{\frac{\pi}{4}}\left(\int\limits_0^{\frac{n}{\cos(\theta)}}\dfrac{r}{1+r}\mathrm{d}r\right)\mathrm{d}\theta=2\cdot\int\limits_0^{\frac{\pi}{4}}\left(r-\log(1+r)\right)\Big|_0^{\frac{n}{\cos(\theta)}}\mathrm{d}\theta.$$
  The evaluation of the integrand yields:
  $$\frac{n}{\cos(\theta)}-\log(1+\frac{n}{\cos(\theta)}).$$

For completeness's sake, the limits for the other part are 0 and $\frac{n}{\sin(\theta)}$, making it the same as this one, as is obvious in cartesian coordinates by a symmetry of the function.
Now we need but to complete the evaluation of that integral. For that, I need the primitive of $\frac{1}{\cos\theta}$:
\begin{align*}
\int\frac{1}{\cos\theta}d\theta={}&\frac{\theta}{\cos\theta}-\int\frac{\theta}{-\cos^2\theta}(-\sin\theta) d\theta=\frac{\theta}{\cos\theta}-\int\theta\frac{d}{d\theta}(\tan\theta)\sin\theta d\theta={} \\
{}={}&\frac{\theta}{\cos\theta}-\theta\tan\theta\sin\theta+\int\theta\tan\theta(\theta\cos\theta+\sin\theta)d\theta={} \\
{}={}&\frac{\theta}{\cos\theta}-\theta\tan\theta\sin\theta+\int\left(\theta\sin\theta+\tan\theta\sin\theta\right)d\theta.
\end{align*}
The first term in that last integral integrates to $-\theta\cos\theta+\sin\theta$. The second term integrates to $-\tan\theta\cos\theta+\int\frac{\cos\theta}{\cos^2\theta}d\theta$, oh dear, all this work for nothing… If I substitute $t=\cos\theta$, the integral becomes $\frac{1}{t\sqrt{1-t^2}}dt$. We substitute $u=\sqrt{1-t^2}$ and get $t=\sqrt{1-u^2}$ and $du=-\frac{t}{\sqrt{1-t^2}}dt$ or $dt=-\frac{u}{\sqrt{1-u^2}}du$, so the new integral is $-\frac{1}{u\sqrt{1-u^2}}\frac{u}{\sqrt{1-u^2}}du=-\frac{1}{1-u^2}du=-\frac12(\frac{1}{1-u}+\frac{1}{1+u})du$, so this evaluates to $-\frac12\log(1-u)-\frac12\log(1+u)=-\frac12\log(1-u^2)$, and going back to $t$ this is $-\frac12\log t^2$, and back into $\theta$ we get $-\frac12\log(\cos^2\theta)=-\log(\cos\theta)$. Which is evidently wrong, so what am I doing wrong? I won't even attempt to set the other term straight.
