How do you find the integral of #sqrt(x^2-25)/x#?

1 Answer
Mar 31, 2018

#int sqrt(x^2-25)/x dx =sqrt(x^2-25) - arctan(sqrt(x^2-25)/5)+C#

Explanation:

The function is defined for #abs x >=5#. Restrict for the moment to #x > 5# and substitute:

#x = 5 sect#

#dx = 5sect tant dt#

with #t in [0,pi/2)#.

Then:

#int sqrt(x^2-25)/x dx = 5 int (sqrt(25sec^2t -25) sect tant )/(5sect)dt#

#int sqrt(x^2-25)/x dx = 5 int sqrt(sec^2t -1) tant dt#

Use now the trigonometric identity:

#sec^2-1 = tan^2t#

and as for # in [0,pi/2)# the tangent is positive:

#sqrt(sec^2-1) = tant#

so:

#int sqrt(x^2-25)/x dx = 5 int tan^2t dt#

using the same identity again:

#int sqrt(x^2-25)/x dx = 5 int (sec^2t-1) dt#

and for the linearity of the integral:

#int sqrt(x^2-25)/x dx = 5 int sec^2tdt -5int dt#

#int sqrt(x^2-25)/x dx = 5tant-5t +C#

To undo the substitution note that:

#tant = sqrt(sec^2t-1) = sqrt((x/5)^2-1) = 1/5 sqrt(x^2-25)#

and then:

#t = arctan(sqrt(x^2-25)/5)#

So:

#int sqrt(x^2-25)/x dx =sqrt(x^2-25) - arctan(sqrt(x^2-25)/5)+C#

By differentiating both sides we can verify that the solution extends also to #x < -5#.