Calka z n
: 14 sty 2008, o 17:33
\(\displaystyle{ \int x^n \ln x dx}\)
\(\displaystyle{ lnx=u\iff \frac{dx}{x}=du}\)
\(\displaystyle{ x^ndx=dv\iff \frac{x^{n+1}}{n+1}=v}\)
\(\displaystyle{ \frac{lnx\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\int \frac{x^{n+1}}{x}dx=
\frac{lnx\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\int x^n dx=
\frac{lnx\cdot x^{n+1}}{n+1}-\frac{x^{n+1}}{(n+1)^2}+C}\)