What is the antiderivative of y=x/square root(x^2-9) ?

We'll substitute x^2 - 9 =
t

We'll differentiate both
sides:

2xdx = dt

xdx =
dt/2

We'll re-write the integral in
t:

Int xdx/sqrt (x^2 - 9) = (1/2)*Int dt/sqrt
t

(1/2)*Int dt/sqrt t = (1/2)*t^(-1/2 + 1)/(1 - 1/2) +
C

(1/2)*Int dt/sqrt t = (1/2)*t^(1/2)/(1/2) +
C

(1/2)*Int dt/sqrt t = t^(1/2) +
C

(1/2)*Int dt/sqrt t = sqrt t +
C

We'll substitute t by the original expression x^2 -
9:

The antiderivative of f(x) is: Int
xdx/sqrt (x^2 - 9) = sqrt(x^2 - 9) + C