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

To find \( \int y \, dx \) for the given functions, we'll tackle them one by one. 1. For \( y = \frac{\ln x}{x} \): We can solve the integral by using integration by parts. Let's set: - \( u = \ln x \) which implies \( du = \frac{1}{x} dx \) - \( dv = \frac{1}{x} dx \) which implies \( v = \ln x \) Using integration by parts, \( \int u \, dv = uv - \int v \, du \), we get: \[ \int \frac{\ln x}{x} \, dx = \int u \, dv = \ln x \cdot \ln x - \int \ln x \cdot \frac{1}{x} \, dx \] We recognize that the remaining integral is the same as the original, resulting in: \[ I = \int \frac{\ln x}{x} \, dx \] From this, we can rearrange to find: \[ 2I = (\ln x)^2 \implies I = \frac{1}{2} (\ln x)^2 + C \] Thus, \( \int \frac{\ln x}{x} \, dx = \frac{1}{2} (\ln x)^2 + C \). 2. For \( y = \frac{\sin^{-1}(4x)}{\sqrt{1 - 16x^2}} \): This integral can be evaluated by noticing that the derivative of \( \sin^{-1}(4x) \) is \( \frac{4}{\sqrt{1 - (4x)^2}} = \frac{4}{\sqrt{1 - 16x^2}} \). Therefore, we can set \( u = \sin^{-1}(4x) \), which implies: \[ du = \frac{4}{\sqrt{1 - 16x^2}} dx \implies dx = \frac{\sqrt{1 - 16x^2}}{4} du \] Substituting back into the integral gives us: \[ \int \frac{\sin^{-1}(4x)}{\sqrt{1 - 16x^2}} dx = \frac{1}{4} \int u \, du \] Thus: \[ = \frac{1}{4} \cdot \frac{u^2}{2} + C = \frac{1}{8} (\sin^{-1}(4x))^2 + C \] In conclusion, we have: 1. \( \int \frac{\ln x}{x} \, dx = \frac{1}{2} (\ln x)^2 + C \) 2. \( \int \frac{\sin^{-1}(4x)}{\sqrt{1 - 16x^2}} \, dx = \frac{1}{8} (\sin^{-1}(4x))^2 + C \)