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

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Answer 1 · 2015-10-26T16:00:30

Using the trigonometric substitution technique of integration we eventually get
#int(sqrt(x^2-16))/xdx=sqrt(x^2-16)-4sec^(-1)(x/4)+C#

I will use the trigonometric substitution technique of integration to solve this integral.

Let #u=4 sec theta#
#therefore du = 4 sectheta tantheta d theta#

Therefore the original integral becomes :

#int (sqrt(4^2sec^2 theta - 4^2))/(4sectheta)*4sec theta tan theta d theta#

#=int sqrt(4^2(sec^2 theta - 1))/(4sectheta)*4sec theta tan theta d theta#

#=int 4 tan^2 theta d theta#

#=4 (tan theta - theta) + C#

#=(sqrt(x^2-16))-4sec^(-1)(x/4)+C#