2 proste calki do sprawdzenia
: 5 gru 2008, o 21:43
1)
\(\displaystyle{ \int_{}^{} (\frac{1-x}{x})^{2}dx= \int_{}^{}\frac{1-2x+x^{2}}{x^{2}}dx= \int_{}^{} x^{-2}dx-2 \int_{}^{} x^{-1}+ \int_{}^{} dx= \\ =-x^{-1}-2ln|x|+x+c}\)
2)
\(\displaystyle{ \int_{}^{} xsinxdx=}\)
u=x v'=sinx
u'=1 v=-cosx
\(\displaystyle{ =xcosx+ \int_{}^{} cosxdx= xcox+sinx}\)
Edit: poprawione, thx
\(\displaystyle{ \int_{}^{} (\frac{1-x}{x})^{2}dx= \int_{}^{}\frac{1-2x+x^{2}}{x^{2}}dx= \int_{}^{} x^{-2}dx-2 \int_{}^{} x^{-1}+ \int_{}^{} dx= \\ =-x^{-1}-2ln|x|+x+c}\)
2)
\(\displaystyle{ \int_{}^{} xsinxdx=}\)
u=x v'=sinx
u'=1 v=-cosx
\(\displaystyle{ =xcosx+ \int_{}^{} cosxdx= xcox+sinx}\)
Edit: poprawione, thx