Descompunem fractia in fractii simple:
[tex]\displaystyle\frac{2x+3}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}[/tex]
Aducem la acelasi numitor:
[tex]2x+3=A(x+2)+B(x+1)\\2x+3=Ax+2A+Bx+B\\2x+3=(A+B)x+2A+B\\\left \{ {{A+B=2\Rightarrow B=2-A} \atop {2A+B=3}} \right. \\2A+2-A=3\Rightarrow A=1\Rightarrow B=1\\\displaystyle\frac{2x+3}{(x+1)(x+2)}=\frac1{x+1}+\frac1{x+2}\\\displaystyle\int\limits_0^1\frac{2x+3}{(x+1)(x+2)}dx=\int\limits_0^1\frac1{x+1}+\frac1{x+2}dx=\int\limits_0^1\frac1{x+1}dx+\int\limits_0^1\frac1{x+2}dx=\ln(x+1)\big|_0^1+\ln(x+2)\big|_0^1=\ln2-\ln1+\ln3-\ln2=\ln3[/tex]
Sau puteam observa ca numaratorul e de fapt suma numitorilor:
[tex]\displaystyle\int\limits_0^1\frac{2x+3}{(x+1)(x+2)}dx=\int\limits_0^1\frac{(x+1)+(x+2)}{(x+1)(x+2)}dx=\int\limits_0^1\frac{x+1}{(x+1)(x+2)}+\frac{x+2}{(x+1)(x+2)}dx=\int\limits_0^1\frac{1}{x+2}+\frac{1}{x+1}dx=...[/tex]
