\( \int \ln ( x ) \operatorname { arctg } x d x \)

To tackle the integral \( \int \ln(x) \operatorname{arctg}(x) \, dx \), we can use integration by parts. Let \( u = \ln(x) \) and \( dv = \operatorname{arctg}(x) \, dx \). Thus, we then need to find \( du \) and \( v \). 1. Differentiate \( u \): \[ du = \frac{1}{x} \, dx \] 2. Integrate \( dv \): To find \( v \), we compute the integral of \( \operatorname{arctg}(x) \). The integral can be found using integration techniques or a table of integrals, yielding: \[ v = x \operatorname{arctg}(x) - \frac{1}{2} \ln(1+x^2) \] Now applying integration by parts: \[ \int u \, dv = uv - \int v \, du, \] we have: \[ \int \ln(x) \operatorname{arctg}(x) \, dx = \ln(x) \left( x \operatorname{arctg}(x) - \frac{1}{2} \ln(1+x^2) \right) - \int \left( x \operatorname{arctg}(x) - \frac{1}{2} \ln(1+x^2) \right) \frac{1}{x} \, dx. \] This may simplify your work, but the integral depends on careful handling of logarithmic functions and their relationships with inverse trigonometric functions. Don't forget to apply limits appropriately if evaluating a definite integral. Happy integrating!