Question

de ce este rezolvata asa integrala aceasta? este ceva formula ?​

Answer 1

$\displaystyle I = \int \dfrac{x}{\sqrt{x^2+1}}\, dx \\ \\ \sqrt{x^2+4} = t \Rightarrow (\sqrt{x^2+4})'\, dx = t'\, dt \Rightarrow \dfrac{2x}{2\sqrt{x^2+4}}\, dx = dt \Rightarrow \\ \\ \Rightarrow \dfrac{x}{\sqrt{x^2+4}}\, dx = dt \\\\ \Rightarrow I = \int dt = t+C= \sqrt{x^2+4}+C\\ \\\text{Fiindca: }\\ \\\int \dfrac{x}{\sqrt{x^2+4}}\, dx=\int (\sqrt{x^2+4})'\, dx =\sqrt{x^2+4}+C$