Integral Exercise 25

We aim to evaluate the following integral:

$$ \int \frac{1}{x \sqrt{1 - \log^2 x}} \ dx $$

To proceed, we use a differential substitution based on \( \log x \):

$$ d(\log x) = \frac{1}{x} \ dx $$

By the invariance property of equations, we multiply both sides by \( x \):

$$ x \cdot d(\log x) = \frac{1}{x} \cdot dx \cdot x $$

This simplifies to an expression for \( dx \):

$$ dx = x \cdot d(\log x) $$

Now we substitute dx = x · d(log x) into the integral:

$$ \int \frac{1}{x \sqrt{1 - \log^2 x}} \cdot \left( x \cdot d(\log x) \right) $$

The \( x \) terms cancel out, leaving:

$$ \int \frac{1}{\sqrt{1 - \log^2 x}} \cdot d(\log x) $$

Let’s now make the substitution \( t = \log x \):

$$ \int \frac{1}{\sqrt{1 - t^2}} \ dt $$

This is a standard form whose antiderivative is the arcsine function:

$$ \int \frac{1}{\sqrt{1 - t^2}} \ dt = \arcsin(t) + c $$

Substituting back \( t = \log x \), we obtain:

$$ \arcsin(\log x) + c $$

Therefore, the final result is:

$$ \int \frac{1}{x \sqrt{1 - \log^2 x}} \ dx = \arcsin(\log x) + c $$

And that completes the solution.