Całka nieoznaczona
: 9 lut 2009, o 12:21
Jak rozwiazac \(\displaystyle{ \int \frac{ \frac{1}{3}x+ \frac{4}{3} }{ x^{2}-x+1 }}\)
ernest180 pisze:Jak rozwiazac \(\displaystyle{ \int \frac{ \frac{1}{3}x+ \frac{4}{3} }{ x^{2}-x+1 }}\)
\(\displaystyle{ \int \frac{ \frac{1}{6} (x ^{2}-x+1)'+ \frac{9}{6} }{x ^{2}-x+1 }= \frac{1}{6}ln|x ^{2}-x+1|+ \frac{9}{6} \int \frac{dx}{(x- \frac{1}{2} ) ^{2}+ (\frac{ \sqrt{3} }{2}) ^{2} }=}\)
\(\displaystyle{ \frac{1}{6}ln|x ^{2}-x+1|+ \frac{3}{ \sqrt{3} }arctg( \frac{2x-1}{ \sqrt{3} }) +C}\)