Question

In=$\int\limits^0_1 { \frac{x^{n} +1}{x+1} } \, dx$ atunci $I_{n} + I_{n+1}$ = ?

Answer 1

[tex] I_{n}+ I_{n+1} = \int\limits^0_1 { \frac{ x^{n}+1+x^{n+1}+1 }{x+1} } \, dx = \int\limits^0_1 { x^{n} } \, dx + \int\limits^0_1 { \frac{2}{x+1} } \, dx = \frac{ x^{n+1} }{n+1} [/tex]+ln(x+1) de la 1 la 0, inlocuind se obtine  : = -1/(n+1) -ln2