[tex] \int \sec hx[/tex](adsbygoogle = window.adsbygoogle || []).push({});

I solve it in this way:

[tex] \int \arccos hx[/tex]

[tex] \int \ln (x^2 + \sqrt{x^2 -1}) dx [/tex]

Then, I substitute [tex] u = \ln (x + \sqrt{x^2 + 1})[/tex]

then I get

[tex] x\ln(x + \sqrt{x^2 + 1}) - \int x/\sqrt{x^2 + 1}dx[/tex]

and then I substitute v = x^2 + 1

[tex] x\ln(x + \sqrt{x^2 + 1}) -1/2 \int v^(1/2) dv [/tex]

and finally I get the answer,

[tex] x\ln(x + \sqrt{x^2 + 1}) -\sqrt{x^2 + 1} + C [/tex]

Am I right?

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# Homework Help: How to find the integral of the secant function

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