Question

Question edited :
I only don't know how to integrate x/(x^2+x+1)
Please help with this

Answer 1

http://www.wolframalpha.com/input/?i=x%2F%28x^3-1%29

Answer 2

\[
\frac x{x^3-1}=\frac { A x + B}{x^2+ x+1}+ \frac C {x-1}

\]

Answer 3

yeah but it does't work , cant even integrate a nimber like that the first term Ax+B

Answer 4

Multiply the two sides by x-1, you get
\[
\frac {x(x-1)}{x^3-1}=\frac {( A x + B)(x-1)}{x^2+ x+1}+ \frac {C(x-1)} {x-1}\\

\frac {x}{x^2+ x+1}=\frac {( A x + B)(x-1)}{x^2+ x+1}+ C
\]
Put x=1
you get\[
\frac 1 3= 0 + C\]
We have C =1/3

Answer 5

yeah c=1/3 b=-1/3 and A=C

Answer 6

Now multiply the two sides by x and let x goes to infinity
\[

\frac {x^2}{x^3-1}=\frac { A x^2 + Bx}{x^2+ x+1}+ \frac x {3(x-1) }\\
0=A + \frac 13\\
A=-\frac 13
\]

Answer 7

Now
\[
\frac x{x^3-1}=\frac { A x + B}{x^2+ x+1}+ \frac1 {3(x-1)}
\\
x=0
\\
0 =\frac { + B}{+1}+ \frac1 {3(0-1)}\\
B=\frac 13
\]

Answer 8

I think eldi has already done the partial fractions as shown on the third line of the problem.

Answer 9

I did it by what we call the cover-up method
\[
\frac{x}{x^3-1}=\frac{1-x}{3 \left(x^2+x+1\right)}+\frac{1}{3 (x-1)}
\]

Answer 10

this is the same answer i got

Answer 11

but i can't integratate this fraction still

Answer 12

\[\frac{1-x}{3 \left(x^2+x+1\right)} = \frac{ 1 }{ 3 } * [ \frac{3}{2 \left(x^2+x+1\right)} - \frac{2x+1}{2 \left(x^2+x+1\right)} ]\]

Answer 13

i am not sure which is right but the part (x^2 +x+1 ) i still can not integrate that ???

Answer 14

a general rule,
\(\\ \text{ To integrate }\quad \huge \frac{ax+b}{\sqrt{px^2+qx+r}} or \frac{ax+b}{px^2+qx+r} \\ \large express \quad ax+b=A\frac{d}{dx}(px^2+qx+r)+B \\ \text{then separate D.}
\)

Answer 15

**separate the Denominator

Answer 16

for 1/(x^2+x+1) try to complete the square in the denominator

Answer 17

i am trying completing the square ,

Answer 18

you separated (1-x)/(denom) = 1/ denom - x/ denom ??
don't do that, do what ranga proposed.
did you get what he did ?

Answer 19

\(\frac{1-x}{3 \left(x^2+x+1\right)} = \frac{ 1 }{ 3 } * [ \frac{3}{2 \left(x^2+x+1\right)} - \frac{2x+1}{2 \left(x^2+x+1\right)} ]\)

Answer 20

I changed the question on top, that is the only thing i don't get from the problem

Answer 21

ok then.
since d/dx (x^2+x+1) is 2x+1

we write "x" in numerator as
(1/2) (2x)
= (1/2) (2x+1 - 1)
= (1/2) (2x+1) - 1/2

then separate the denominators.

Answer 22

Thank you very much, you see I am have a test on two days and I have been studying all day couldn't get around this one than you again

Answer 23

oh, so you could solve it entirely?!
ask if you atill have any doubt in any step :)

Answer 24

no no it's ok i get it now ,,but i have never seen that type of manipulation before the one you wrote

Answer 25

ohh, with such types, linear / quadratic or linear/ sqrt (quadratic)
such manipulation is quite common :)
good luck for your test!

Answer 26

Here is a similar solved problem. There is an integral sign missing in the line before last.
This problem was generated by http://www.saab.org/calculus.cgi and selecting Techniques of Integrations.