What is the indefinite integral of 1/x(ln^2x + 9) x > 0 ?

To solve the indefinite integral of the given function,
we'll use the substitution technique.

We'll put ln x =
t.

We'll differentiate both
sides:

dx/x = dt

We'll
re-write the function in t and we'll calculate the indefinite
integral:

Int f(x)dx = Int dt/(t^2 +
9)

We'll use the identity:

Int
dx/(x^2 + a^2) = [arctan (x/a)]/a + C

Comparing, we'll
get:

Int dt/(t^2 + 3^2) = [arctan (t/3)]/3 +
C

The indefinite integral of f(x) is: Int
f(x)dx = [arctan (ln x/3)]/3 + C