Question

Integrala de la 0 la 1 din x/x^2+3x+2 dx.

Answer 1

$\boxed{ Metoda I: }$
$\int\limits^1_0 { \dfrac{x}{x^2+3x+2} } \, dx = \dfrac{1}{2}\cdot \int\limits^1_0 { \dfrac{2x}{x^2+3x+2} } \, dx = \dfrac{1}{2}\cdot \int\limits^1_0 { \dfrac{2x+3-3}{x^2+3x+2} } \, dx = \\ \\ = \dfrac{1}{2}\cdot \int\limits^1_0 { \Big(\dfrac{2x+3}{x^2+3x+2} } - \dfrac{3}{x^2+3x+2} \Big)} \, dx = \\ \\ = \dfrac{1}{2}\cdot \int\limits^1_0 { \dfrac{2x+3}{x^2+3x+2} } \, dx - \dfrac{1}{2}\cdot \int\limits^1_0 { \dfrac{3}{x^2+3x+2} } \, dx =$

$=\int\limits^1_0 { \dfrac{(x^2+3x+2)'}{x^2+3x+2} } \, dx - \dfrac{3}{2}\cdot \int\limits^1_0 { \dfrac{1}{x^2+3x+2} } \, dx = \\ \\ =\Big(ln(x^2+3x+2)\Big)\Big|_0^1 - \dfrac{3}{2} \cdot \int\limits^1_0 { \dfrac{1}{x^2+2\cdot \dfrac{3}{2} x+ \Big(\dfrac{3}{2}\Big) ^2-\Big(\dfrac{3}{2}\Big) ^2 +2} } \, dx= \\ \\ =ln(1+3+2)-ln(0+0+2) - \dfrac{3}{2} \cdot\int\limits^1_0 { \dfrac{1}{\Big(x+ \dfrac{3}{2}\Big)^2- \dfrac{1}{4} } } \, dx =$
$=ln6-ln2 - \dfrac{3}{2}\cdot \int\limits^1_0 { \dfrac{\Big(x+ \dfrac{3}{2}\Big)'}{\Big(x+ \dfrac{3}{2}\Big)^2- \Big(\dfrac{1}{2}\Big)^2 } } \, dx = \\ \\ = ln \dfrac{6}{2} - \dfrac{3}{2}\cdot \Big(\dfrac{1}{2\cdot \dfrac{1}{2} } \cdot ln\Big| \dfrac{x+ \dfrac{3}{2}- \dfrac{1}{2} }{x+ \dfrac{3}{2}+ \dfrac{1}{2}}\Big| \Big) \Big|_0^1 = \\ \\ =ln3 - \dfrac{3}{2} \cdot \Big(ln\Big| \dfrac{x+ 1 }{x+ 2}\Big| \Big) \Big|_0^1 =$


$=ln3 - \dfrac{3}{2} \cdot \Big(ln\Big( \dfrac{1+ 1 }{1+ 2}\Big) - ln \Big(\dfrac{0+ 1 }{0+ 2}\Big) = 3 - \dfrac{3}{2} \cdot\Big(ln \dfrac{2}{3} - ln \dfrac{1}{2} \Big) = \\ \\ =ln3 - \dfrac{3}{2} \cdot ln \Big( \dfrac{2}{3}\cdot \dfrac{2}{1}\Big) = ln 3 - \dfrac{3}{2} \cdot ln \dfrac{4}{3}$

$\boxed{ Metoda 2: } \\ \\ \dfrac{x}{x^2+3x+2} = ? \quad ( este o functie rationala, o impartim in 2 fractii ) \\ \\ x^2+3x+2 = 0, \quad \Delta = 9-8 = 1 \Rightarrow x_1_2 = \dfrac{-3\pm1}{2} \Rightarrow \\ \Rightarrow x_1 = -1,\quad x_2 = -2 \\ \\ \Rightarrow \dfrac{x}{x^2+3x+2} = \dfrac{x}{(x+1)(x+2)} = \dfrac{^{(x+2)}^\slash A}{x+1}+ \dfrac{^{(x+1)}^\slash B}{x+2} = \\ \\ = \dfrac{Ax+2A+Bx+B}{(x+1)(x+2)} = \dfrac{(A+B)\cdot x + 2A + B}{(x+1)(x+2)}$

$\dfrac{x}{(x+1)(x+2)} =\dfrac{(A+B)\cdot x + 2A + B}{(x+1)(x+2)} \Rightarrow \\ \Rightarrow \Big\{A+B = 1 \quad si \quad 2A+B = 0\Big\} ( scadem) \Rightarrow A \ = -1 \\ -2+B = 0 \Rightarrow B = 2 \\ \\ \Rightarrow \dfrac{x}{(x+1)(x+2)} = \dfrac{-1}{x+1} + \dfrac{2}{x+2} \\ \\ \Rightarrow \int\limits^1_0 { \dfrac{x}{x^2+3x+2} } \, dx = \int\limits^1_0 { \dfrac{x}{(x+1)(x+2)} } \, dx = \int\limits^1_0 {\Big( \dfrac{-1}{x+1} + \dfrac{2}{x+2} \Big)} \, dx =$

$=\int\limits^1_0 { \dfrac{-1}{x+1}} \, dx + \int\limits^1_0 {\dfrac{2}{x+2} } \, dx = - \int\limits^1_0 { \dfrac{1}{x+1}} \, dx + 2 \int\limits^1_0 {\dfrac{1}{x+2} } \, dx = \\ \\ =\Big(-ln(x+1)+2ln(x+2)\Big)\Big|_0^1 = \\ \\ =-ln(1+1)+2ln(1+2)+ln(0+1)-2ln(0+2) = \\ =-ln2+2ln3+0-2ln2 = -3ln2+2ln3 = -ln2^3+ln3^2 = \\ =ln \Big(\dfrac{3^2}{2^3} \Big) = ln \dfrac{9}{8}$