calculus - How to get the same answer for integral $intfrac{dz}{sqrt{1+z^2}}$ as in textbook

I have some answer:
$$\int\frac{dz}{\sqrt{1+z^2}}=\ln ( z + \sqrt {1 + z^2})+C$$
But I can't get the same expression. So here what I've done and got stuck. How can I get the same answer?

$$\begin{align} & \int\frac{dz}{\sqrt{1+z^2}}=\left\{z=\tan t, \ dz=\frac{dt}{\cos^2 t} \right\} = \int\frac{dt}{\cos^2 t\sqrt{1+\tan^2 t}} \\[10pt] = {} & \left\{\sqrt{1+\tan^2 t}=\frac{1}{\cos t}\right\}=\int\frac{\cos t \ dt}{\cos^2 t} \\[10pt] = {} & \left\{\cos t \ dt= d\sin t, \cos^2 t=1-\sin^2 t\right\}=\int\frac{d\sin t}{1-\sin^2 t} \\[10pt] = {} &\{\sin t=u\}=\int\frac{d u}{1-u^2}=-\int\frac{du}{u^2-1}=-\frac{1}{2} \int\left(\frac{1}{u-1}-\frac{1}{1+u}\right)\,du \\[10pt] = {} &-\frac{1}{2}(\ln |u-1|-\ln|u+1|)+C \end{align}$$

From here I don't understand how to get the answer above. Need help

Answer

You're almost there. Note that with $u=\sin(t)$ and $z=\tan(t)$, we have

$$\begin{align} -\frac12\left(\log(|u-1|)-\log(|u+1|) \right)&=\frac12\log\left(\left|\frac{1+\sin(t)}{1-\sin(t)}\right|\right)\\\\ &=\frac12\log\left(\left|\frac{(1+\sin(t))^2}{1-\sin^2(t)}\right|\right)\\\\ &=\log\left(\left|\frac{1+\sin(t)}{\cos(t)}\right|\right)\\\\ &=\log(|\sec(t)+\tan(t)|)\\\\ &=\log(|z+\sqrt{1+z^2}|) \end{align}$$

as was to be shown!