Question

Va rog sa ma ajutati! Dau coroana+ 70.puncte

Answer 1

f(x) = $\frac{1}{x^{2} +5x+6}$

a) $f(x) = \frac{1}{x^{2} +5x+6} = \frac{1}{x^{2} +3x+2x+6} = \frac{1}{x(x+3)+2(x+3) } = \frac{1}{(x+3)(x+2)}$

Separi fractia in 2 parti: 1/(x+3)(x+2) = A/(x+3) + B/(x+2) si vei inmulti ambele parti ale ecuatiei cu (x+3)*(x+2)

1 = (x+2)A + (x+3)B

1 = Ax+Bx+2A+3B

1 = (A+B)x+(2A+3B)

Ca sa dea 1, unul dintre termeni trebuie sa fie 1 si -1

Daca A = -1 si B = 1 ⇒(-1+1)x + (-2+3) = 0+1 = 1 Adevarat

Din $f(x) = \frac{1}{(x+3)(x+2)} = \frac{A}{x+3} + \frac{B}{x+2} = \frac{-1}{x+3} + \frac{1}{x+2} = \frac{1}{x+2} - \frac{1}{x+3}$

b) $\int\limits^1_0 {\frac{1}{x^{2} +5x+6} } \, dx = \int\limits^1_0 {\frac{1}{x+2} -\frac{1}{x+3} } \, dx = \int\limits^1_0 {\frac{1}{x+2} - \int\limits^1_0 {\frac{1}{x+3} = ln(|x+2|) -ln(|x+3|) ^1_0 =$

$= (ln2 - ln4) - (ln2- ln3) = -ln4 +ln3 = ln(\frac{3}{4} )$

c)

$\int\limits^1_2 {\frac{1}{x(x+1)} } \, dx = \int\limits^1_2 \frac{1}{x} - \frac{1}{x+1}dx = (ln|x| - ln(|x+1|))^1_2 = (ln1 - ln2) - (ln2-ln3) = ln1 +ln3 = ln(1*3) = ln3$