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 :

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

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

$=int 4 tan^2 theta d theta$

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

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

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