Question

Nu imi dau seama cum sa calculez INTEGRALA DIN 1 SUPRA 5-Xpatrat

Answer 1

$\it \dfrac{1}{5-x^2}=\dfrac{1}{(\sqrt5)^2-x^2}= \dfrac{1}{(\sqrt5-x)(\sqrt5+x)}\\ \\ \\ \dfrac{1}{(\sqrt5-x)(\sqrt5+x)} = \dfrac{a}{\sqrt5-x} +\dfrac{b}{\sqrt5+x} = \dfrac{a\sqrt5+ax+b\sqrt5-bx}{(\sqrt5-x)(\sqrt5+x)}=\\ \\ \\ =\dfrac{(a-b)x+\sqrt5(a+b)}{(\sqrt5-x)(\sqrt5+x)}\Rightarrow \begin{cases} \it a-b=0\Rightarrow a=b\ \ \ (*)\\ \\\it \sqrt5(a+b)=1\stackrel{(*)}{\Longrightarrow} \sqrt5\cdot2a=1\Rightarrow a=\dfrac{1}{2\sqrt5}\end{cases}$

$\it Deci\ \dfrac{1}{5-x^2}=\dfrac{1}{2\sqrt5}\left(\dfrac{1}{\sqrt5-x}+\dfrac{1}{\sqrt5+x}\right)= \dfrac{1}{2\sqrt5}\left(\dfrac{-1}{x-\sqrt5}+\dfrac{1}{x+\sqrt5}\right)\\ \\ \\=\dfrac{1}{2\sqrt5}\left(\dfrac{1}{x+\sqrtt5}-\dfrac{1}{x-\sqrt5}\right)\\ \\ \\\int \dfrac{1}{5-x^2} dx = \int\dfrac{1}{2\sqrt5} \left(\dfrac{1}{x+\sqrtt5}-\dfrac{1}{x-\sqrt5}\right)dx=\\ \\ \\=\dfrac{1}{2\sqrt5}\left(\int\dfrac{1}{x+\sqrt5}dx-\int\dfrac{1}{x-\sqrt5}dx\right) =$

$\it = \dfrac{1}{2\sqrt5}(ln|x+\sqrt5|-\ ln|x-\sqrt5|) +\mathcal{C}=\dfrac{1}{2\sqrt5}ln\Big{|}\dfrac{x+\sqrt5}{x-\sqrt5}} \Big{|}+\mathcal{C}$