Saved Bookmarks
| 1. |
Find:\( \int \frac{\left(x^{2}-1\right)}{x^{2}+x+1} d x \) |
|
Answer» \(\int\frac{(x^2-1)dx}{x^2+x+1}\) Let x2 - 1 = A(x2 + x + 1) + B(2x + 1) + C ⇒ x2 - 1 = Ax2 + (A + 2B)x + A + B + C \(\therefore\) A = 1, A + 2B = 0 and A + B + C = -1 ⇒ B = -1/2 and C = -1 - A - B = -1 - 1 + 1/2 = -2 + 1/2 = -3/2 \(\therefore\) x2 - 1 = (x2 + x + 1) - 1/2 (2x + 1) - 3/2 ⇒ \(\frac{x^2-1}{x^2+x+1}=1-\frac12\times\frac{2x+1}{x^2+x+1}dx\) - \(\frac32\int\frac{1}{x^2-x+1}dx\) = x - 1/2 log|x2 + x + 1| - 3/2 \(\int\cfrac{1}{(x+1/2)^2+(\sqrt3/2)^2}dx+C\) = x - 1/2 log|x2 + x + 1| - 3/2 x \(\frac1{\sqrt3/2}\) tan-1(\(\frac{x+1/2}{\sqrt2/3}\)) + C = x - 1/2 log |x2 + x + 1| - √3 tan-1(\(\frac{2x+1}{\sqrt3}+C\)) |
|