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[tex] \int\limits^e_1 \frac{x}{x^2+1} \, dx [/tex] ???

Răspuns :

u = x² +1 
du = 2x·dx
amplificam integrala cu 2 
avem u' /u 
I= 1/2·ln( x² +1 ) = 1/2  [ ln (e²+1 ) - ln2]= 1/2 [ ln ( e²+1) /2 ]
t=x^2+1
dt=2xdx
Aflam F(x)= integrala de la x/ x^2+1 dx= 1/2 integrala din dt/t= 1/2 lnt+ C= 1/2 lnx^2+1  +C
Aflam:
F(e)= 1/2 +lne^2+1
F(1)= 1/2+ln2
F(e)-F(1)= 1/2(lne^2 +1 -ln2)= 1/2 (lne^2+1/2).