Question

integral 1/(x^2-x+1) dx
Steps?

Answer 1

complete the square maybe

Answer 2

Trig sub for the win
Need to write bottom as the difference of two squares

Answer 3

and let me say "squares"

Answer 4

the bottom is NOT a diff of 2 squares tho is it?

Answer 5

x^2 - 2*(1/2) x + (1/4) +(1-1/4)

=> (x-1/2)^2 + (sqrt(3)/2)^2

Answer 6

\[\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\]

Answer 7

lol...
\[x^2-x+1=x^2-x+(\frac{1}{2})^2+1-(\frac{1}{2})^2=(x-\frac{1}{2})^2+1-\frac{1}{4}=(x-\frac{1}{2})^2+\frac{3}{4}\]
\[=(x-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2\]
lol yes Or i mean the sum of two squares
pardon my earlier french

Answer 8

or one of the two depending on what you gots

Answer 9

how does the (1/2)^2 come into the equation?

Answer 10

\[\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\frac{4/3}{4/3}=\frac{4/3}{\frac{4}{3}(x-\frac{1}{2})^2+1} \]

i dont spose theres a way to get this into the tan-1 stuff ...

Answer 11

partial decomp in the complexes i spose

Answer 12

Thanks!

Answer 13

yw, hope it helps