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[tex] \int \frac{\sqrt{x^2-3}}{x} dx[/tex]

Using trig substitution

[tex]c^2=a^2+b^2[/tex]

[tex]a = \sqrt{c^2-b^2}[/tex]

[tex]∴ c = x, b = \sqrt{3}[/tex]

Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..

[tex] \int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx[/tex]

And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.