Presupunem ca am ajuns la raspunsul arctg(√(x²-1))+c dar nu il gasim in variantele de raspuns. (Si acesta este tot raspuns corect)
Incercam schimbarea de variabila 1/x in schimb, deoarece 1/x se afla in celelalte variante.
[tex]f:(-\infty, -1)\to \mathbb{R},\quad f(x) = \dfrac{1}{x\sqrt{x^2-1}}\\ \\ \displaystyle I =\int \dfrac{1}{x\sqrt{x^2-1}}\,dx\\ \\ \dfrac{1}{x} = t \Rightarrow x = \dfrac{1}{t}\Rightarrow dx = -\dfrac{1}{t^2}\, dt\\ \\ I = -\int \dfrac{t}{\sqrt{\dfrac{1}{t^2}-1}}\dfrac{1}{t^2}\, dt = -\int \dfrac{1}{t\sqrt{\dfrac{1-t^2}{t^2}}}\, dt =-\int \dfrac{1}{t\cdot \dfrac{1}{|t|}\sqrt{1-t^2}}\, dt =[/tex]
[tex]\displaystyle x < -1 \Rightarrow \dfrac{1}{x} < -1 \Rightarrow t<-1\Rightarrow |t| = -t \\ \\ = -\int \dfrac{1}{t\cdot \dfrac{1}{-t}\sqrt{1-t^2}}\,dt= -\int \dfrac{1}{-\sqrt{1-t^2}}\, dt = \int \dfrac{1}{\sqrt{1-t^2}}\, dt = \\ \\ = \arcsin(t)+c = \arcsin\dfrac{1}{x}+c\\ \\ \Rightarrow \boxed{B}- \text{corect}[/tex]
