What is the indefinite integral of 1/x(ln^2x + 9) x > 0 ?
We need to find the integral of f(x) = [1/(x(ln x)^2 + 9)]
Let ln x = y
=> dy/dx = 1/x
=> dy = dx / x
Int [f(x) dx] = Int [ 1/x*((ln x)^2 + 9) dx]
=> Int [ 1/ (y^2 + 9) dy]
=> [arc tan (y / 3)] / 3 + C
substitute y = ln x
=> [arc tan (ln x / 3)] / 3 + C
The required integral is [arc tan (ln x / 3)] / 3 + C
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