∫ dx / ((x – 3) ln4(x – 3))
Табличный интеграл ∫ u–4du=u–4+1/(–4+1)+C=–(1/3)·(1/u3)+C u=ln(x–3) du=dx/(x–3) О т в е т. (–1/3)·(1/(ln(x–3))3) + C