Question

Интеграл (x-3)log2 xdx?

Answer 1

∫log₂(x-3)dx=

u=log₂(x-3) => du=ln2dx/(x-3)
dv=dx => v=x

=x•log₂(x-3)-ln2•∫((x-3+3)/(x-3))dx=x•log₂(x-3)-ln2•∫(1+3/(x-3))dx=
=x•log₂(x-3)-ln2•(x+3•ln|x-3|)+C