Before anything, I'd like to clarify that I have no background in calculus (I'm still in school). I only try to learn calculus as a hobby. Please be gentle.  I'm trying to evaluate the aforementioned integral but the answer I got doesn't seem to match with 3 sources different sources that I found online. My process was to substitute in $x = \sec u, \ dx = \tan u\sec u\, du$. This led me to the answer $\tan (\mathrm{arcsec}\ x) - \mathrm{arcsec}\ x$. However after trying to check my answer I saw that  says it's $\arctan \left(\frac 1{\sqrt{x^2-1}}\right) + \sqrt{x^2-1}$,  says it's $-\arctan (\sqrt{x^2-1}) +  \sqrt{x^2-1}$, and this  says it's $-\mathrm{arcsec} \,x + \sqrt{x^2-1}$.  I plotted all four results on a . Symbolab and WolframAlpha only differ by a constant so I assume that both of them have the correct result whereas the video and I have the wrong ones. Also it can be seen that all the results are the same on the positive side of the x axis but things break down on the negative side.  My final questions are:    Where did the video creator and I go wrong? What is happening on the negative x axis?   EDIT: It seems this post might be a duplicate of . I hope this post doesn't get removed because I believe what I asked is a bit different from what was asked there. EDIT 2: Upon further inspection, I guess it's reasonable for this post to be removed.
My try, using $x = \sec(u)$ substitution:  $$ \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &amp;=&amp; \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &amp;=&amp; \int \tan^2(u) \mathrm{d}u \\ &amp;=&amp; \tan(u) - u + C \\ &amp;=&amp; \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} $$  However, according to , the answer should be: $$ \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C $$ When I derive this last answer I don't get back the integrand, but rather: $$ \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} $$  I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.  Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
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