Indefinite integral of arcsec(x)

[PWiz]

Just for fun, I tried this rather trivial problem, but I think I went wrong somewhere:
$$\int arcsec(x) \ dx$$
Let $arcsec(x)=y$ . Then $x=sec \ y$, or $y=arccos(\frac 1{x})$
So the problem becomes $$\int arccos(\frac 1 {x}) \ dx$$
Let $\frac 1 {x} = cos \ u$ , so that $dx = secu \ tanu \ du$.
So the questions simplifies to $\int u \ secu \ tanu \ du$. If I integrate by parts, then this equals to
$$u \int secu \ tanu \ du - \int (\int secu \ tanu \ du) du$$
Since $\int secu \ tanu \ du = secu$, I get $u \ secu - \int secu \ du$
$\int secu \ du = ln(\sqrt{\frac{1+sinu}{1-sinu}})$
Substituting the Xs, I get $$x \ arcsec(x) - ln(\sqrt{\frac{1+\sqrt{1-\frac{1}{x^2}}}{1-\sqrt{1-\frac{1}{x^2}}}}) + c= x \ arcsec(x) - ln(\sqrt{\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}}) + c $$
where $c$ is an arbitrary constant. I checked my answer with Wolfram Alpha here:
http://integrals.wolfram.com/index.jsp?expr=arcsec(x)&random=false
As you can clearly see, the answer is different from mine. In which step did I goof?

 

[Raghav Gupta]

, I get $$x \ arcsec(x) - ln(\sqrt{\frac{1+\sqrt{1-\frac{1}{x^2}}}{1-\sqrt{1-\frac{1}{x^2}}}}) + c= x \ arcsec(x) - ln(\sqrt{\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}}) + c $$
where $c$ is an arbitrary constant. I checked my answer with Wolfram Alpha here:
http://integrals.wolfram.com/index.jsp?expr=arcsec(x)&random=false
As you can clearly see, the answer is different from mine. In which step did I goof?

Your all steps are correct, you have not goofed.
Wolfram's answer as well as your answer is correct. It's the way of manipulating.
Rearrange some terms in wolfram' answer and in yours. You will see both are same.

Edit [ Add] indefinite integration is an area where you get answers which are not matching by someone's else. The constant of integration has a big role here.

 

[PWiz]

Okay, thanks!