Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 4

Use integration by parts with $u=(\ln x{)}^{n+1}$ and $dv=dx$. Since $du=(n+1)(\ln x{)}^{n}dx/x$, this yieldsAs an identity between indefinite integrals, the formula is understood up to an additive constant.

Here $I_0=x$ up to a constant. Substitution of $n=0$ in the integration by parts recurrence givesIndeed, its derivative is $\ln x+1-1=\ln x$.