Question

Salut, am si eu nevoie de ajutor la ex. 941 (explicatii pas cu pas )

Answer 1

$\displaystyle I =\int\dfrac{1}{x\sqrt{x^2+1}}\, dx =\int \dfrac{1}{x\sqrt{x^2\Big(1+\dfrac{1}{x^2}}\Big)}\, dx = \int \dfrac{1}{x\cdot |x|\cdot \sqrt{1+\dfrac{1}{x^2}}}\, dx =\\ \\ x>0\Rightarrow |x| = x \\ \\ I = \int \dfrac{1}{x^2\sqrt{1+\dfrac{1}{x^2}}}\, dx =-\int x\cdot \dfrac{1}{-x^3\sqrt{1+\dfrac{1}{x^2}}}\, dx\\ \\ \sqrt{1+\dfrac{1}{x^2}} = t \Rightarrow -\dfrac{\dfrac{2}{x^3}}{2\sqrt{1+\dfrac{1}{x^2}}}\, dx = dt \Rightarrow -\dfrac{1}{x^3\sqrt{1+\dfrac{1}{x^2}}}\, dx = dt$

$\displaystyle \sqrt{1+\dfrac{1}{x^2}} = t \Rightarrow 1+\dfrac{1}{x^2} = t^2 \Rightarrow \dfrac{1}{x} = \sqrt{t^2-1}\Rightarrow x = \dfrac{1}{\sqrt{t^2-1}}\\ \\ I = \int\dfrac{1}{\sqrt{t^2-1}}\, dt = -\ln(t+\sqrt{t^2-1})+C = \\ \\ =-\ln\Big(\sqrt{1+\dfrac{1}{x^2}}+\sqrt{1+\dfrac{1}{x^2}-1}}\Big)+C = - \ln\Big(\sqrt{1+\dfrac{1}{x^2}}+\sqrt{\dfrac{1}{x^2}}}\Big)+C = \\ \\ = -\ln\Big(\dfrac{1}{x}+\sqrt{1+\dfrac{1}{x^2}}\Big)+C\\ \\ \Rightarrow c)\text{ corect}$