Mathematics

This integral pairs a logarithm with a power function—a classic setup for integration by parts. The key insight is to let $u = \ln t$ (which gets simpler when differentiated) and $dv = t^{-3}\,dt$ (which is easy to integrate). After applying the parts formula, you'll get a product term to evaluate at the bounds, plus a remaining integral that's straightforward to handle.

This integral pairs a logarithm with a power function—a classic setup for integration by parts. The key insight is to let $u = \ln t$ (which gets simpler when differentiated) and $dv = t^{-3}\,dt$ (which is easy to integrate). After applying the parts formula, you'll get a product term to evaluate at the bounds, plus a remaining integral that's straightforward to handle.