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How do you find the Integral of #dx/sqrt(x^2+16)#?

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Mar 30, 2015

#int (dx)/ sqrt(x^2+16) dx = int 1/(4sqrt((x/4)^2+1)) dx#

#=int 1/(sqrt((x/4)^2+1))( 1/4)dx #

#u = x/4#, #du=1/4 dx# and #int 1/sqrt(u^2+1) du=sinh^(-1) u +C#

So
#int (dx)/ sqrt(x^2+16) dx = sinh^(-1) (x/4) +C#

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