Question

integrate (((2/3)-(2x/3))/(x^2+x+1))

Answer 1

just sub divide then cross multiply and add the outliers to x.

Answer 2

understand?

Answer 3

\[\int\limits_{}^{}\frac{ \frac{ 2 }{ 3 }-\frac{ 2x }{ 3 }}{ x^{2}+x+1}\\]

Answer 4

no quite

Answer 5

oh

Answer 6

first you need to factories quadratically

Answer 7

and use calculus to find the sectum

Answer 8

factorise quadratically?

Answer 9

find sectum?

Answer 10

yes..
u subdivide to x then factorise the quadratic equation on the right.

Answer 11

split them into 2 different equations
you have

[(2/3) / (x^2+x+1)] - [2x/3 / (x^2+x+1)]
simplify
[2x^2 + 2x + 1]/3 - [2x^3 + 2x^2 + 2x] / 3
simplify more

2/3x^2 + 2/3x + 1/3 - 2/3x^3 - 2/3x^2 - 2/3x
and your integrating that which is

2/9x^3 + 1/3x^2 + 1/3x - 1/6x^4 - 2/9x^3 -1/3x^2 + c
simplify
-x^4/6 + x/3 + C is your answer

Answer 12

then you find the sectum

Answer 13

and cross multiply.. understood

Answer 14

?
0

Answer 15

btw jayz you made a huge mistake

Answer 16

you forgot to add x to the 3xsquared when you were subdividing..

Answer 17

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

Answer 18

wrong
you failed at subdividing again you

Answer 19

uhm, that should be \[\large -\frac{ 1 }{ 3 }\frac{ 2x+1 }{ x^2+x+1 }+\frac{ 1 }{ (x+1/2)^2+(\sqrt{3/4})^2 }\]

Answer 20

the first term is the ln function, the second term is arctan function

Answer 21

the answer is