The provided text contains several math problems. We will solve problem "I" which is an integral: $I = \int (2x+1) \frac{ln(x)}{x} dx$.

1. Problem Description

The provided text contains several math problems. We will solve problem "I" which is an integral:

$I = \int (2x+1) \frac{ln(x)}{x} dx$.

2. Solution Steps

Let's solve the integral $I = \int (2x+1) \frac{ln(x)}{x} dx$. We can rewrite the integral as:

$I = \int (2 + \frac{1}{x}) ln(x) dx = \int 2ln(x) dx + \int \frac{ln(x)}{x} dx$.

Let's solve the first integral: $\int 2ln(x) dx$. Using integration by parts, let $u=ln(x)$, $dv=2dx$. Then $du = \frac{1}{x}dx$ and $v=2x$.

$\int 2ln(x) dx = 2xln(x) - \int 2x \frac{1}{x} dx = 2xln(x) - \int 2dx = 2xln(x) - 2x + C_1$.

Now, let's solve the second integral: $\int \frac{ln(x)}{x} dx$. Let $u = ln(x)$, then $du = \frac{1}{x} dx$.

So, $\int \frac{ln(x)}{x} dx = \int u du = \frac{u^2}{2} + C_2 = \frac{(ln(x))^2}{2} + C_2$.

Combining the two results:

$I = 2xln(x) - 2x + \frac{(ln(x))^2}{2} + C$, where $C=C_1 + C_2$.

3. Final Answer

$I = 2xln(x) - 2x + \frac{(ln(x))^2}{2} + C$.