Let's begin by solving the integral shown in question number 37 from the image:

$\int \frac{\ln x}{2x} \, dx$

We can solve this integral using integration by parts. The integration by parts formula is:

$\int u \, dv = uv - \int v \, du$

In this case, we let:

• $u = \ln x$, so $du = \frac{1}{x} \, dx$
• $dv = \frac{1}{2x} \, dx$, so $v = \frac{1}{2} \ln x$

Now applying the formula:

$\int \frac{\ln x}{2x} \, dx = \frac{1}{2} \left( \ln x \cdot \ln x - \int \frac{1}{x} \cdot \ln x \, dx \right)$

Simplifying the integral, we get:

$\frac{1}{2} \left( \ln^2 x - \int \frac{1}{x} \ln x \, dx \right)$

The result gives a complete simplification for the equation.