Răspuns:
2
Explicație pas cu pas:
[tex]I=\int\limits^1_0 {\frac{x+a}{x+1}} \, dx =\int\limits^1_0 {\frac{x+1+a-1}{x+1}} \, dx=\int\limits^1_0 ({\frac{x+1}{x+1}+\frac{a-1}{x+1})} \, dx=\int\limits^1_0 {\frac{x+1}{x+1}} \, dx+\int\limits^1_0 {\frac{a-1}{x+1}} \, dx=\int\limits^1_0 {1} \, dx+(a-1)\int\limits^1_0 {\frac{1}{x+1}} \, dx=x|^1_0+(a-1)ln(x+1)|^1_0=1-0+(a-1)(ln2-ln1)=1+ln2*(a-1)[/tex]
Dar, stim ca rezultatul integralei este: [tex]1+ln2[/tex].
Rezolvam ecuatia:
[tex]1+ln2*(a-1)=1+ln2\\ln2*(a-1)=ln2\\ln2*(a-1)-ln2=0\\ln2*(a-1-1)=0\\ln2*(a-2)=0\\a-2=0\\a=2[/tex]
