[tex]\int\limits\frac{x}{x^2-2x+3}dx[/tex]
[tex]\int\limits\frac{x}{x^2-2x+3} dx=\int\limits\frac{x-1+1}{x^2-2x+3} dx=\int\limits\frac{x-1}{x^2-2x+3} dx +\int\limits\frac{1}{x^2-2x+3}dx=[/tex]
Pentru prima integrala: u(x)=x²-2x+3
u'(x)=2x-2=2(x-1)
[tex]\frac{1}{2} \int\limits\frac{(x^2-2x+3)'}{(x^2-2x+3}dx=\frac{1}{2} ln(x^2-2x+3)[/tex]
Pentru a doua integrala:
[tex]\int\limits\frac{1}{x^2-2x+3}dx=\int\limits\frac{1}{(x-1)^2+(\sqrt{2})^2} }dx=\frac{1}{\sqrt{2} } arctg\frac{(x-1)}{\sqrt{2} }[/tex]
Numitorul este sub forma unei ecuatii de gradul al doilea.
Delta da negativa, asa ca am scris sub forma canonica a ecuatiei de gr.2: a(x+b/2a)+ (-Δ)/(4a)
Deci, integrala da:
[tex]\frac{1}{2} ln(x^2-2x+3)+\frac{1}{\sqrt{2}}arctg\frac{(x-1)}{\sqrt{2}} +C[/tex]
