How do you integrate #sqrt(x^2-9)/(x)dx#?

1 Answer
Feb 17, 2015

To integrate this function we will use a trigonometric substitution.

Let #costheta=3/x#

Therefore, #sectheta=x/3# and #x =3sectheta#

Differentiate #x=3sectheta#

#dx=3secthetatanthetad theta#

Make the substitution into the integral

#intsqrt((3sectheta)^2-9)/(3sectheta)3secthetatanthetad theta#

#intsqrt(9sec^2theta-9)tantheta d theta#

#intsqrt(9(sec^2theta-1))tanthetad theta#

#intsqrt(9)sqrt(tan^2theta)tanthetad theta#

#3inttan^2thetad theta#

#3intsec^2theta-1d theta#

Now integrating we get

#3(tantheta-theta)#

Now define #theta# and #tantheta # in terms of #x# as follows

#theta=arctan(sqrt(x^2-9)/3) #

#tantheta = sqrt(x^2-9)/3 #

now back substitute

#3(sqrt(x^2-9)/3-arctan(sqrt(x^2-9)/3))+C #

Distributing the #3# we will have

#sqrt(x^2-9)-3arctan(sqrt(x^2-9)/3)+C # FINAL ANSWER