Calka
: 19 cze 2019, 00:58
\(\int_{-1}^{2} (sgn(x^3-x ) +\frac{3x-8}{x^2-6x+8}) dx\)
\(\int_{-1}^{2} (sgn(x^3-x ) dx=\int_{-1}^{2} (sgn \left[ (x+1)x(x-1)\right] dx= \int_{-1}^{0}1dx+ \int_{0}^{1}(-1)dx+ \int_{1}^{2}1dx=1-1+1=1\)
\(\int_{-1}^{2} \frac{3x-8}{x^2-6x+8} dx =\int_{-1}^{2} \frac{3x-8}{(x-2)(x-4)} dx =\int_{-1}^{2} \left( \frac{1}{x-1}+ \frac{2}{x-4} \right) dx =...\)