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

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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**