If $\Large \int \frac {dx}{(x+1)(x-2)} \normalsize = A\; log(x+1)+B\; log(x-2)+c,then$ $$(a)\;A+B=0 \qquad (b)\;AB=- \qquad(c)AB=\frac{1}{9}\qquad(d)\;AB=-9$$

Share

Please log in or register to add a comment.