Question

Sa se determine integralele nedefinite:
1. ∫$\frac{2x^3-x^4}{ \sqrt{x} }$ dx
2. ∫[tex](x \sqrt[5]{x} - \sqrt[3]{x^2} -x \sqrt[4]{x}) dx
[/tex]
3.∫$\frac{x \sqrt[3]{x} +2x^2 \sqrt[4]{x^2} }{ \sqrt{x} } dx$
4. ∫$\frac{ \sqrt{x^2+4 }-1}{x^2+4} dx$

Answer 1

1 
(2x³-x^4)/√x=2*x^(3-1/2)-x^(4-1/2)=            √x=x^1/2
=2x^(5/2)-x^7/2
F(x)= primitiva
F(x)=∫2x^(5/2)dx-∫x^7/2dx=
Se  aplica  formula  ∫x^αdx=1/(α+1)*x^(α+1)  α∈R
F(x)=2/(5/2+1)*x^(5/2+1)-1/(7/2+1)*(x^(7/2+1) +c=
4/7*x^(7/2)-2/9*x^(9/2)+c
_____________________
Ex2
Se introduc  introduc  numerele  sub  radical  tinand  con t ca  x≥0
Vom  avea
F(x)=∫x^(5+1)/5dx-∫x^(2/3)dx_∫x^(4+1)/4dx=
∫x^6/5dx-∫x^2/3dx-∫x^5/4dx=
Se  aplica  formula  de  la  ex  1
1/(6/5+1)*x^(6/5+1)-1/(2/3+1)*x^(2/3+1)-1/(5/4+1)*x^(5/4+1)=
5/11*x^11/5-3/5*x^2/5-4/9*x^9/4+c