<s>[INST] Verify the integral $\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$ I'm stuck solving the integral
$$\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$$
This is what I got so far
\begin{align}
\int_{1}^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}} - u} \,du &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{u - u\cdot(1 + \sqrt{u^2 + 1})} \,du \\
 &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{-u\cdot \sqrt{u^2 + 1}} \,du = -\bigg( \underbrace{\int_1^z \frac{1}{u\sqrt{u^2 + 1}} \,du}_{(*)} + \int_1^z \frac{1}{u}\,du\bigg) \\
 &= - (*) - \ln(z)
\end{align}
Calculation of $(*)$ yields
\begin{align*}
 (*) &= \int_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{u}{u^2  \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{\varphi'(u)}{\varphi(u)^2 - 1} \,du = \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u^2 -1} \,du \\
 &= \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u-1} \,du - \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u+1} \,du = \frac{1}{2} \bigg[\ln(u-1) - \ln(u+1)\bigg]_{\sqrt{2}}^{\sqrt{z^2 + 1}} \\
 &= \frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right)
\end{align*}
Thus we have the result
\begin{align*}
 &-\frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right) - \ln(z) \\
 &= -\frac{1}{2} \left(\ln(\frac{(\sqrt{z^2 + 1} - 1)^2}{(\sqrt{z^2 + 1} + 1) \cdot (\sqrt{z^2 + 1} - 1)}) - \ln(\frac{(\sqrt{2} - 1)^2}{(\sqrt{2} + 1)\cdot(\sqrt{2} - 1)})\right) - \ln(z) \\
 &= -\frac{1}{2} \left(\ln(\frac{z^2 + 1 - 2\sqrt{z^2 + 1} + 1}{z^2}) - \ln(\frac{2 - 2\sqrt{2} + 1}{2 - 1})\right) - \frac{1}{2}\ln(z^2) \\
 &= -\frac{1}{2} \left(\ln(z^2 - 2\sqrt{z^2 + 1} + 2) - \ln(3 + 2\sqrt{2})\right) \\
 &= \frac{1}{2} \ln(3 + 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)
\end{align*}
The correct solution is given by 
$$\ln(\frac{\sqrt{z^2 + 1} + 1}{z^2(1 + \sqrt{2})})$$
I'm not able to simplify my solution to the given solution. Did I integrate wrong or is there a way to verify the solution?
Edit: I think, that I made an mistake somewhere, since the solution should equal zero for $z = 1$ but $1$ is not root of my solution..
Edit 2: I found the mistake. I switched "+" and "-" in the line of the second last equality. My question remains how to simplify my solution to the given solution.. My solution then looks like this
$$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
 [/INST] $$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
$$=\frac{1}{2}{ln(\sqrt{2}-1)^2}-\frac{1}{2}{ln(\sqrt{z^2+1}-1)^2}$$
$$=\frac{1}{2}({ln \frac{(\sqrt{2}-1)^2}{(\sqrt{z^2+1}-1)^2})}$$
$$={ln \frac{(\sqrt{2}-1)}{(\sqrt{z^2+1}-1)}}$$
$$={ln \frac{(\sqrt{z^2+1}+1)}{z^2(\sqrt{2}+1)}}$$
</s>