∫ (x5+∛(x2)+(1/∛x))dx=
= ∫ x5dx + ∫ x2/3dx+ ∫ x–1/3dx=
=(x6/6) +x5/3/(5/3) +x2/3/(2/3)+C=
= (x6/6) +(3/5)·x5/3 +(3/2)·x2/3+C=
= (x6/6) +(3/5)·x·∛(x2) +(3/2)·∛(x2)+C
м)
∫ (7x–2)4dx= ∫ (7x–2)4·(1/7)d(7x–2)=
=(1/7) ∫ u4du=(1/7)(u5/5) + C= (1/35)·(7x–2)5 + C
u=7x–2
du=(7x–2)`=7dx ⇒ dx=(1/7)d(7x–2)
т)
=2 ∫ dx/(1+x2)–3 ∫ dx/√1–x2=
=2arctgx – 3 arcsinx + C