Question

The integral of (x^2 - x)/(x^2 + x +1) from 0 to 7

Answer 1

Do I use long division?

Answer 2

yes

Answer 3

so I have 1-(1/x^2 +x +1); correct?

Answer 4

1-(2x+1)/(x^2+x+1

Answer 5

Yu got it?

Answer 6

no....

Answer 7

Try your division again
X^2/X^2=1
so 1(x^2+X+1)=X^2+X+1
(X^2-X)-(x^2+X+1)=-2X-1=-(2X+1)

Answer 8

\[\int\limits_{0}^{7}\frac{x^2-x}{x^2+x+1}dx=\int\limits_{0}^{7}\frac{x^2+x+1-2x-1}{x^2+x+1}dx=\int\limits_{0}^{7}\left(1-\frac{2x+1}{x^2+x+1}\right)dx=\]
\[=\int\limits_{0}^{7}dx-\int\limits_{0}^{7}\frac{d(x^2+x+1)}{x^2+x+1}=x|^7_0-\ln{(x^2+x+1)}|^7_0=7-\ln{57}\]

Answer 9

natural log, instead of log....that's where i went wrong
Thanks!