Determine the indefinite integral of y=x/square root(x^2+9), using substitution.

We'll substitute the radicand x^2+9 by
t.

x^2+9 = t

We'll
differentiate both sides:

d(x^2+9)/dx =
dt

2xdx = dt => xdx =
dt/2

We'll solve the indefinite integral in
t:

Int x dx/sqrt(x^2+9) = Int dt/2sqrt t = (1/2)*Int
dt/t^1/2

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

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

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

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

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

The indefinite integral of the given
function is: Int xdx/sqrt(x^2+9) = sqrt(x^2 + 9) +
C.