4

$$I = \int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$$

Substituting for $e^x$,

$$I = \int {u^2 - 1 \over u^4 + u^2 + 1} du = \int { u^4 + u^2 + 1 + - 2 - u^4 \over u^4 + u^2 + 1} du = u - \int {u^4 + 2 \over u^4 + u^2 + 1} du $$

Now I don't know anything I can do to last integral except partial fraction decomposition but I am pretty sure that $u^4 + u^2 + 1$ does not have any factors in real numbers.

Is this integral computable on real numbers ? How do I compute it ?

  • 2
    $u^4+u^2+1$, like any real polynomial with no real roots, can be factored into irreducible quadratic factors. In this case they should be straightforward to calculate. (Also you lost a minus sign.) – Ian Apr 15 '17 at 01:37
  • @Ian How can I find those factors ? I literally plugged every number from 1 to 10. –  Apr 15 '17 at 01:40
  • 1
    It is easy to factor $u^4+u^2+1$ over $\mathbb{C}$ because it is just $v^2+v+1$ where $v=u^2$, so each root is one of the square roots of the roots of $v^2+v+1$. Then you can multiply out the linear factors corresponding to conjugate roots to get irreducible quadratic factors. (This is not the only way to do it of course.) – Ian Apr 15 '17 at 01:41
  • @Ian With any polynomial do you mean polynomial with 4th degree or just any nth degree polynomial ? –  Apr 15 '17 at 10:26
  • Technically any real polynomial will do, but if the degree is higher than 4 then you may not have a closed form for the roots and thus for the factorization. – Ian Apr 15 '17 at 10:43
  • @Ian Thank you very much for this. –  Apr 15 '17 at 10:55

2 Answers2

5

Observe that $$u^4 + u^2 + 1 = (u^4 + 2u^2 + 1) - u^2 = (u^2 + 1)^2 - u^2 = (u^2 + u + 1)(u^2 - u + 1).$$ From this, we try a partial fraction decomposition of the form $$\frac{Au + B}{u^2 - u + 1} + \frac{Cu + D}{u^2 + u + 1} = \frac{u^2 - 1}{u^4 + u^2 + 1},$$ and after multiplying out and comparing like coefficients of $u$, we have $$\begin{align*} A + C &= 0 \\ A+B-C+D &= 1 \\ A+B+C-D &= 0 \\ B+D &= -1. \end{align*}$$ From here, we get $$B = D = -1/2,$$ and $$A = 1, C = -1.$$ Consequently, the integrand becomes $$\frac{1}{2}\left( \frac{2u-1}{u^2-u+1} - \frac{2u+1}{u^2+u+1}\right),$$ and the rest is straightforward. As a bonus, you even find that the numerator of each term is the derivative of the respective denominator. You can't get much nicer than that.

heropup
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4

We don't actually need tedious Partial Fraction Decomposition as

$$\dfrac{u^2-1}{u^4+Au^2+1}=\dfrac{1-\dfrac1{u^2}}{u^2+A+\dfrac1{u^2}}$$ where $A$ is an arbitrary constant.

Now as $\displaystyle\int\left(1-\dfrac1{u^2}\right)du=u+\dfrac1u,$

write $\displaystyle u^2+A+\dfrac1{u^2}=\left(u+\dfrac1u\right)^2+A-2$ and set $u+\dfrac1u=v$

lab bhattacharjee
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