1
$\begingroup$

Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$

Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$

could some help me how to solve,thanks

$\endgroup$
  • $\begingroup$ I would suggest factorising...Partial Fractions and then integrating $\endgroup$ – user35508 Mar 20 '17 at 5:10
  • $\begingroup$ Hint: the denominator factors as $(x-1)(x^2+x-1)(x^3+x+1)$ $\endgroup$ – Grant B. Mar 20 '17 at 5:11
5
$\begingroup$

$\dfrac{2x - \frac{1}{x^2}}{x^4 -x^2 +2x -2 -\frac{1}{x} + \frac{1}{x^2}}$

We want to write the denominator in $x^2 + \frac{1}{x}$ since it is derivative in the nominator Reordering

$$x^4 +2x +\frac{1}{x^2}-2 -x^2 - \frac{1}{x} = \left(x^2 +\frac{1}{x}\right)^2 - 2 -(x^2 + \frac{1}{x}) $$

Let $ u = x^2 + \frac{1}{x^2}$

$\endgroup$
  • $\begingroup$ Beautiful solution! +1 $\endgroup$ – Michael Rozenberg Mar 20 '17 at 13:13
3
$\begingroup$

$$\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx=\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{2x+1}{x^2+x-1}-\frac{3x^2+1}{x^3+x+1}\right)dx=$$ $$=\frac{1}{3}\left(\ln|x-1|+\ln|x^2+x-1|-\ln|x^3+x+1|\right)+C$$

$\endgroup$
1
$\begingroup$

As Grant B. commented $$x^6-x^4+2x^3-2x^2-x+1=(x-1)(x^2+x-1)(x^3+x+1)$$ Then, Using partial fraction decomposition,

$$\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}=\frac 13 \left(\frac{2 x+1}{ x^2+x-1}-\frac{3 x^2+1}{ x^3+x+1}+\frac{1}{ x-1}\right)$$ where you can notice that each numerator is the derivative of each denominator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.