How do you evaluate the integral #int dx/sqrt(a^2+x^2)#?

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Answer 1 · 2017-02-24T18:40:23

#= sinh^(-1) (x/a) + C#

Because #cosh^2 z - sinh^2 z = 1#, I would go with the hyperbolic sub #x = a sinh y, dx = a cosh y \ dy#

The integral becomes:

#int 1/sqrt(a^2+a^2 sinh^2 y)a cosh y \ dy#

#= int 1/(a cosh y)*a cosh y \ dy#

#= y + C#

#= sinh^(-1) (x/a) + C#

And because #sinh^(-1) z = ln ( z + sqrt(z^2 + 1))#

#= ln ( x/a + sqrt((x/a)^2 + 1)) + C#