a) faci substitutia 1+e^x=y=>e^xdx=dy
integrala devine
∫dy/ y=lny+C
revii la substitutie
F(x)=ln(1+e^x)+c
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d)F(x)=∫lnxdx/x
faci substitutia lnx=y dx/x=dy integrala devine
F(y)=∫dy=y+C
Revii la substitutie
F(x)=lnx+c
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I=∫cosxdx/(1+sin²x) x∈[0,π/2]
Faci substitutia sinx=t =.>.cosxdx=dt
Schimbi limitele de integrare
x=0 sin0=t=0
x=π/2 sinπ/2=t=1
I=∫dt/(1+t²) t∈[0,1]
I=arctgt /o↑1=arctg1-arctg 0=π/4-0=π/4
