Wyznaczyć całki
: 09 sie 2010, 13:36
a) \(\int_{}^{} (x-2x)^2 dx\)
b)\(\int_{}^{} x^2/(x^3 + 3)^2 dx\)
c)\(\int_{}^{} lnx * 1/x^2 dx\)
b)\(\int_{}^{} x^2/(x^3 + 3)^2 dx\)
c)\(\int_{}^{} lnx * 1/x^2 dx\)
a)
\(\int_{}^{} (x-2x)^2dx= \int_{}^{} (-x)^2dx= \int_{}^{} x^2dx= \frac{1}{3}x^3+C\)
c)
\(\int_{}^{} \frac{lnx}{x^2}dx\)
Stosuję wzór na całkowanie przez części : \(\int_{}^{} f \cdot g'dx=f \cdot g- \int_{}^{} f' \cdot gdx\)
\(g'= \frac{-1}{x^2}\;\; \Rightarrow \;\;g= \frac{1}{x}
f=lnx\;\;\; \Rightarrow \;\;\;f'= \frac{1}{x}\)
\(-1 \int_{}^{}lnx \cdot ( \frac{-1}{x^2})dx=-1[ \frac{lnx}{x}- \int_{}^{} \frac{1}{x} \cdot \frac{1}{x}dx]=
=-1[ \frac{lnx}{x}- \int_{}^{} x^(^-^2)dx]=-1[ \frac{lnx}{x}+ \frac{1}{x}]=- \frac{lnx}{x}- \frac{1}{x}+C\)