Question

cat este integrala ∫√1+x^2 dx ?
radicalul cuprinde si x patrat

Answer 1

$\int\limits { \sqrt{1+ x^{2} } } \, dx= \int\limits{ \frac{1+ x^{2} }{ \sqrt{1+ x^{2} } } } \, dx= \int\limits{ \frac{1}{ \sqrt{1+ x^{2} } } } \, dx+ \int\limits{x \frac{2x}{2 \sqrt{1+ x^{2} } } } \, dx$ , A doua interala se face prin parti:
f=x    ⇒ f'=1
g'=$\frac{2x}2 \sqrt{1+ x^{2} }$;g=$\sqrt{1+ x^{2} }$
Deci  I=$ln(x+ \sqrt{1+ x^{2} })+x \sqrt{1+ x^{2} }-I$ ⇒ I=$\frac{1}{2}[ln(x+ \sqrt{1+ x^{2} })+x \sqrt{1+ x^{2} }]+C$.