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\int \operatorname{arcsec} x\,dx\,

Use integration by parts with u=\operatorname{arcsec} x\, and dv=dx\,. Then, for x>1\,, du={\frac  {dx}{x{\sqrt  {x^{2}-1}}}}\, and v=x\,. So

\int \operatorname{arcsec} x\,dx =\int u\,dv=uv-\int v\,du
=x\operatorname{arcsec} x-\int {\frac  {x\,dx}{x{\sqrt  {x^{2}-1}}}}
=x\operatorname{arcsec} x-\int {\frac  {dx}{{\sqrt  {x^{2}-1}}}}.

Now use the trigonometric substitution x=\sec \theta \, on the latter integral, with dx=\sec \theta \,\tan \theta \,d\theta . Then

\int {\frac  {dx}{{\sqrt  {x^{2}-1}}}} =\int {\frac  {\sec \theta \,\tan \theta \,d\theta }{{\sqrt  {\sec ^{2}\theta -1}}}}
=\int {\frac  {\sec \theta \,\tan \theta \,d\theta }{{\sqrt  {\tan ^{2}\theta }}}}
=\int \sec \theta \,d\theta =\ln \left|\sec \theta +\tan \theta \right|+C.

As \sec \theta =x\,, drawing a right triangle with hypotenuse of length x and adjacent side of length 1 yields \tan \theta ={\sqrt  {x^{2}-1}}. Putting all of this together then gives the solution

\int \operatorname{arcsec} x\,dx=x\operatorname{arcsec} x-\ln \left(x+{\sqrt  {x^{2}-1}}\right)+C, where the absolute value in the logarithm may be dropped due to the restriction x>1.\,