Question

cat face Integrala din 1 supra x patrat plus 1 ?

Answer 1

sec 0 = secanta

u = x^2 + 1, du/dx = 2x, du/2x = dx

[tex] \int\limits^ {} \frac{1}{1+ x^{2} } \, dx = \frac{1}{u} \frac{du}{2x} [/tex]

[tex]x = tan 0 [/tex]

[tex] \frac{dx}{d0}=sec^{2}0 dx=sec^{2}0 d0 [/tex]

[tex] x^{2} +1 = (tan 0)^{2}+1 = tan^{2}0 + 1 =sec 0 [/tex]

$\int\limits^ {} \frac{1}{x^{2}+1 } \, dx = \int\limits^ {} \frac{1}{sec0}( sec^{2}0 d0) = \int\limits {} sec0 d0\, = ln(tan 0 +sec 0 ) +C =$

$=\ \textgreater \ ln[x+( x^{2} +1)] + C = ln( x^{2} +x+1)$