Question

Buna ziua cum as putea face aceasta integrala

Answer 1

$\displaystyle \\ \\ \int\limits_{2}^4 \dfrac{3}{3x^2+12x+24} \, dx = \int\limits_{2}^4 \dfrac{3}{3(x^2+4x+8)} \, dx = \int\limits_{2}^4 \dfrac{1}{x^2+4x+8} \, dx = \\ \\ =\int\limits_{2}^4 \dfrac{1}{x^2+4x+4+4} \, dx = \int\limits_{2}^4 \dfrac{1}{(x+2)^2+2^2} \, dx = \\ \\ = \int\limits_{2}^4 \dfrac{(x+2)'}{(x+2)^2+2^2} \, dx = \left| \dfrac{1}{2}\cdot \text{arctg}\Big( \dfrac{x+2}{2} \Big)\right|_2^4= \\ \\ = \dfrac{1}{2}\cdot \Big[\text{arctg}\Big( \dfrac{4+2}{2} \Big) - \text{arctg}\Big( \dfrac{2+2}{2} \Big)\Big] =$

$=\dfrac{\text{arctg}( 3 ) - \text{arctg}( 2 )}{2}$

Answer 2

∫​4​​​3x​2​​+12x+24​​3​​dx=​2​∫​4​​​3(x​2​​+4x+8)​​3​​dx=​2​∫​4​​​x​2​​+4x+8​​1​​dx=​​=​2​∫​4​​​x​2​​+4x+4+4​​1​​dx=​2​∫​4​​​(x+2)​2​​+2​2​​​​1​​dx=​​=​2​∫​4​​​(x+2)​2​​+2​2​​​​(x+2)​′​​​​dx=​∣​∣​∣​∣​​​2​​1​​⋅arctg(​2​​x+2​​)​∣​∣​∣​∣​​​2​4​​=​​=​2​​1​​⋅[arctg(​2​​4+2​​)−arctg(​2​​2+2​​)]=​​

=\dfrac{\text{arctg}( 3 ) - \text{arctg}( 2 )}{2}=​2​​arctg(3)−arctg(2)​​