∫ от –1 до 1 (x⁵ / (x + 2)) dx
x5/(x+2)=(x4+2x3+4x2–8x+16+(32/(x+2))
∫ 1–1(x5dx/(x+2))=
=(x5/5)–(x4/2)+(4x3/3)–4x2+16x+32 ln|x+2|)|1–1=
=((1/5)+(1/5))–0 +((4/3)+(4/3))–0 +16·(1+1)+32ln3–32ln1=
=(2/5)+(8/3)+32+32ln3 – 0=
= 35 целых (1/15) + 32ln3