Răspuns:
Explicație pas cu pas:
[tex]\displasytyle\text{Mai intai sa observam ca }2x^2+x-6=(2x-3)(x+2)\\\int \dfrac{3x-1}{2x^2+x-6}dx=\int \dfrac{3x-1}{(2x-3)(x+2)}dx=\int \dfrac{2x-3+x+2}{(2x-3)(x+2)}dx=\\\\ =\int \left(\dfrac{1}{x+2}+\dfrac{1}{2x-3}\right)dx=\int \dfrac{dx}{x+2}+\int \dfrac{dx}{2x-3}=\\ =\boxed{\ln(x+2)+\dfrac{1}{2}\ln(2x-3)+C}[/tex]
