Cam multe
1/ Faci schimbarea de variabila x²=y xdx=dy/2 Integrala devine
∫dy/2*(y²-1)=1/2∫dy/(y²-1)=1/2l*1/2ln l(y-1)/(y+1)l+C=1/4ln l(y-1)/(y+1)l +c
revii la variabila x
F(x)=1/4*lnl (x²-1)/(x²+1)l+c
2)F(x)=∫x²dx/(16-x^6)= -∫dx/(x^6-16)
x³=y 3x²dx=dy x²dx=dy/3
-∫1/3 *dy/(y²-16)=-1/3*∫dy/(y²-(4)²)=-1/3*1/8*lnl (y-4)/(y+4)l+c=-1/24lnl(y-4)/(y+4)l+c
revii la variabila x
F(x)=-1/24*lnl(x³-4)/(x³+4)l+C
3)x²+9=y xdx=dy
FIntegrala devine
∫dy(y=lny+c revii la x
F(x)=ln(x²+9)+c
4)x²=y xdx=dy/2
1/2∫dy/(y²+1)=1/2 arctg y+C REvii la x
F(x)=1/2arctg x²+C
5)faci substitutia x³=y 3x²dx=dy x²dx=dy/3
1/3∫dy/(y²-25)=1/3 ∫dy/(y²-(5)²)=1/3*1/2*5 lnl(y-5)/(y=5)l+c =1/30*ln (y-5)/(y+5)l
F(x)=1/30*lnl(x³-5)/(x³+5)l+C
7)F(x)=∫(x+x³)dx/(1+x^4)=∫xdx/(1+x^4)+∫x³dx/(1+x^4)
F1(x)=∫xdx/(1+x^4)
Schimbare de variabila x²=y xdx=dy/2
F1=∫1/2*dy/(1+y²)=1/2∫dy/(1+y²)=1/2arctgy+c1
Revii la x
F1(x)=1/2arctgx²+c1
F2(x)=∫x³dx/(1+x^4)
1+x^4=y 4x³dx=dy x³dx=dy/4
F2=1/4∫dy/y=1/4lny+c2
F2=1/4ln(1+x^4)+c2
F(x)=1/2arctgx²+1/4ln(1+x^4)+C
