letermine $$ \int y d x $$ in each of the following cases: $$ y=\frac{\ln x}{x} $$ $$ y=\frac{\sin ^{-1} 4 x}{\sqrt{1-16 x^{2}}} $$

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**Case 1**

We need to determine
$$
\int \frac{\ln x}{x}\,dx.
$$
Let
$$
u=\ln x.
$$
Then,
$$
du=\frac{dx}{x}.
$$
Substituting into the integral, we have:
$$
\int \frac{\ln x}{x}\,dx=\int u\,du.
$$
Integrate with respect to $$u$$:
$$
\int u\,du = \frac{u^2}{2}+C.
$$
Returning to the variable $$x$$, we get:
$$
\frac{(\ln x)^2}{2}+C.
$$

---

**Case 2**

We need to determine
$$
\int \frac{\sin^{-1}(4x)}{\sqrt{1-16x^2}}\,dx.
$$
Let
$$
u=\sin^{-1}(4x).
$$
Differentiate $$u$$ with respect to $$x$$. Recall that the derivative of $$\sin^{-1}(z)$$ is $$\frac{1}{\sqrt{1-z^2}}$$ times the derivative of $$z$$; here $$z=4x$$. Thus,
$$
\frac{du}{dx} = \frac{1}{\sqrt{1-(4x)^2}}\cdot 4 = \frac{4}{\sqrt{1-16x^2}}.
$$
This implies
$$
dx=\frac{\sqrt{1-16x^2}}{4}\,du.
$$
Substitute into the original integral:
$$
\int \frac{\sin^{-1}(4x)}{\sqrt{1-16x^2}}\,dx = \int \frac{u}{\sqrt{1-16x^2}}\cdot \frac{\sqrt{1-16x^2}}{4}\,du = \frac{1}{4}\int u\,du.
$$
Now, integrate with respect to $$u$$:
$$
\frac{1}{4}\int u\,du = \frac{1}{4}\cdot \frac{u^2}{2} + C = \frac{u^2}{8}+C.
$$
Returning to the variable $$x$$, we have:
$$
\frac{(\sin^{-1}(4x))^2}{8}+C.
$$
**Case 1:** $$ \int \frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2} + C $$ **Case 2:** $$ \int \frac{\sin^{-1}(4x)}{\sqrt{1-16x^2}}\,dx = \frac{(\sin^{-1}(4x))^2}{8} + C $$

letermine in each of the following cases:

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