The function is defined for abs x >=5|x|≥5. Restrict for the moment to x > 5x>5 and substitute:
x = 5 sectx=5sect
dx = 5sect tant dtdx=5secttantdt
with t in [0,pi/2)t∈[0,π2).
Then:
int sqrt(x^2-25)/x dx = 5 int (sqrt(25sec^2t -25) sect tant )/(5sect)dt∫√x2−25xdx=5∫√25sec2t−25secttant5sectdt
int sqrt(x^2-25)/x dx = 5 int sqrt(sec^2t -1) tant dt∫√x2−25xdx=5∫√sec2t−1tantdt
Use now the trigonometric identity:
sec^2-1 = tan^2tsec2−1=tan2t
and as for in [0,pi/2)∈[0,π2) the tangent is positive:
sqrt(sec^2-1) = tant√sec2−1=tant
so:
int sqrt(x^2-25)/x dx = 5 int tan^2t dt∫√x2−25xdx=5∫tan2tdt
using the same identity again:
int sqrt(x^2-25)/x dx = 5 int (sec^2t-1) dt∫√x2−25xdx=5∫(sec2t−1)dt
and for the linearity of the integral:
int sqrt(x^2-25)/x dx = 5 int sec^2tdt -5int dt∫√x2−25xdx=5∫sec2tdt−5∫dt
int sqrt(x^2-25)/x dx = 5tant-5t +C∫√x2−25xdx=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)tant=√sec2t−1=√(x5)2−1=15√x2−25
and then:
t = arctan(sqrt(x^2-25)/5)t=arctan(√x2−255)
So:
int sqrt(x^2-25)/x dx =sqrt(x^2-25) - arctan(sqrt(x^2-25)/5)+C∫√x2−25xdx=√x2−25−arctan(√x2−255)+C
By differentiating both sides we can verify that the solution extends also to x < -5x<−5.
