Evaluate the integral: I = integral dx/(sqrt(1-x²) arcsin

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{1-x^2}\,\arcsin x}. $$

Explanation of Method

Since $\dfrac{d}{dx}(\arcsin x) = \dfrac{1}{\sqrt{1-x^2}}$, that exact factor multiplies the integrand alongside $1/\arcsin x$. Substituting $u=\arcsin x$ turns the whole integral into the standard logarithmic form $\int \frac{du}{u}$.

Substitute $u=\arcsin x$.
Rewrite the integral in terms of $u$.
Apply the standard logarithmic antiderivative.
Substitute back and add the constant of integration.

Step 1 · Set up the substitution

Let:

$$ u = \arcsin x. $$

Then:

$$ du = \frac{dx}{\sqrt{1-x^2}}. $$

Step 2 · Rewrite the integral

Substituting $u$ and $du$ into the integral:

$$ I = \int \frac{du}{u}. $$

Step 3 · Integrate

This is a standard logarithmic integral:

$$ I = \ln|u| + C. $$

Step 4 · Substitute back

Substituting back $u=\arcsin x$:

$$ I = \ln|\arcsin x| + C. $$

Summary Table

Quantity	Value
Substitution	$u = \arcsin x$
$dx/\sqrt{1-x^2}$ in terms of $du$	$du$
Antiderivative	$\ln\|u\|$

Additional Notes

The domain requires $x\in(-1,1)$ with $x\neq0$ (since $\arcsin0=0$ makes the integrand undefined there).
This follows the same "numerator matches the derivative of an inner function" pattern as the arctangent versions in Problems 107–108, but with $\arcsin x$ in place of $\arctan x$.
The result can be checked by differentiating $\ln|\arcsin x|$ with the chain rule and confirming it returns $\dfrac{1}{\sqrt{1-x^2}\arcsin x}$.

Result

The solution to the integral is:

$$ I = \ln|\arcsin x| + C. $$