Norris Lewis
03/27/2023 · High School
\( \int _{}^{}\frac{1}{x\sqrt {1-\ln (x)^{2}}} d x \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To solve the integral
\[
I = \int \frac{1}{x \sqrt{1 - \ln(x)^2}} \, dx,
\]
we can use a substitution to simplify the integrand. Let's set
\[
u = \ln(x).
\]
Then, the differential \(dx\) can be expressed in terms of \(du\):
\[
dx = e^u \, du.
\]
Also, since \(x = e^u\), we have:
\[
\frac{1}{x} = \frac{1}{e^u}.
\]
Substituting these into the integral, we get:
\[
I = \int \frac{1}{e^u \sqrt{1 - u^2}} e^u \, du = \int \frac{1}{\sqrt{1 - u^2}} \, du.
\]
The integral
\[
\int \frac{1}{\sqrt{1 - u^2}} \, du
\]
is a standard integral that evaluates to
\[
\arcsin(u) + C,
\]
where \(C\) is the constant of integration. Now, substituting back \(u = \ln(x)\), we have:
\[
I = \arcsin(\ln(x)) + C.
\]
Thus, the final result for the integral is
\[
\int \frac{1}{x \sqrt{1 - \ln(x)^2}} \, dx = \arcsin(\ln(x)) + C.
\]
Quick Answer
The integral \( \int \frac{1}{x \sqrt{1 - \ln(x)^2}} \, dx \) simplifies to \( \arcsin(\ln(x)) + C \).
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