1. ∫(x dx)/(5 – 3x2)7
u=5–3x2 du=d(5–3x2)=(5–3x2)`dx=–6xdx xdx=(–1/6)d(5–3x2) ∫ (–1/6)·(5–3x2)–7d(5–3x2)=(–1/6)·(5–3x2)–7+1/(–7+1)= =(1/36)·(1/(5–3x2)6)+C