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