You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Integral dx/(1+x^3)
- Thread starter oooppp
- Start date
Have you tried partial fractions? Give that a try, and Show Us Your Work.I have looked at this and tried all substitution I know, but just can't seem to solve it. Would be very greatful if someone helped me on this.
$\int_0^\infty \frac{dx}{1+x^3}$
View attachment 3346
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Do you know, to start with, that \(\displaystyle x^3+ 1= (x+ 1)(x^2- x+ 1)\)?
Do you know, to start with, that \(\displaystyle x^3+ 1= (x+ 1)(x^2- x+ 1)\)?
yes, that was my attempt, but then I get to integrate \(\displaystyle \frac{BX+C}{(x^2- x+ 1)}\) for some numbers B and C (those I can calculate) which i can't do.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Since \(\displaystyle x^2- x+ 1\) cannot be factored (with real coefficients), complete the square. \(\displaystyle x^2- x+ 1= x^2- x+ 1/4- 1/4+ 1= (x- 1/2)^2+ 3/4\).yes, that was my attempt, but then I get to integrate \(\displaystyle \frac{BX+C}{(x^2- x+ 1)}\) for some numbers B and C (those I can calculate) which i can't do.
Try \(\displaystyle \frac{B(x- 1/2)}{(x- 1/2)^3+ 3/4}+ \frac{C}{(x-1/2)^2+ 3/4}\)
and recall that \(\displaystyle \int \frac{dx}{x^2+ a^2}= arctan(ax)+ C\).
D
Deleted member 4993
Guest
yes, that was my attempt, but then I get to integrate \(\displaystyle \frac{BX+C}{(x^2- x+ 1)}\) for some numbers B and C (those I can calculate) which i can't do.
\(\displaystyle \displaystyle \frac{Bx+C}{x^2 - x + 1} \ = \ \frac{B}{2} * \frac{2x - 1}{x^2 - x + 1} \ + \ \frac{B+2C}{2} * \frac{1}{(x - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}\)
Have you tried partial fractions? Give that a try, and Show Us Your Work.
Here is my attempt
https://www.dropbox.com/s/fsdkk8a189lk54o/2013-10-26 15.38.38.jpg
Here is my attempt. https://www.dropbox.com/s/bghijxpzobb1f5o/2013-10-27 20.26.56.jpg\(\displaystyle \displaystyle \frac{Bx+C}{x^2 - x + 1} \ = \ \frac{B}{2} * \frac{2x - 1}{x^2 - x + 1} \ + \ \frac{B+2C}{2} * \frac{1}{(x - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}\)