корень 6ой степени из (x+4)^5 dx
1) u=x+4 ⇒ du=dx
По формуле
∫х^(α)dx=x^(α+1)/(α+1)
∫u^(5/6)du=u^((5/6)+1)/((5/6)+1)+C=
=(6/11)(x+4)^(11/6)+C
2)u=5-x^2 ⇒ du=-2x dx ⇒ (-1/2)du=xdx
(-1/2)∫u^(1/5)du=(-1/2)u^((1/5)+1)/((1/5)+1)+C=
=(-5/12)(5-x^2)^(6/5)+C
3)sin^2(x/2)=(1-cosx)/2
∫sin^2x(x/2)dx=∫(1-cosx)dx/2=(1/2)∫(1-cosx)=
=(1/2)*(x-sinx) +C
4) u=arctgx ⇒ du=dx/(1+x^2)
∫u^3du=u^4/4 + C=(arctgx)^4/4 + C
5)u=lnx+1 ⇒ du=dx/x
∫udu=u^2/2 + C=(lnx+1)^2/2 + C