1
$\begingroup$

I need to solve integral:

$\int_0^{r(z)} \dfrac{\mathrm{d}z}{r^4(z)-2r^2(z)}$,

where $r(z)=r_i-z(r_i-1)$, where $r_i$ is constant, $z$ is longitudinal coordinate. Boundary condition $r(z)$ is dependent on $z$, so $r$ is not constant and I am not sure how to solve this integral.

If I change boundary immediately in this inegral I will have this shape:

$\int_0^{r(z)} \dfrac{\mathrm{d}z}{a-bz+cz^2-dz^3+ez^4}$, should I solve this integral instead of upper integral and how?

$\endgroup$
  • $\begingroup$ If $z$ is the integration variable, it is bad style, or even possibly hinting at an error in the surrounding computation, to also have it as outer independent variable in the upper end of the integration interval. Please edit to distinguish between these two variables. Or is the integral really $$\int_0^{r(z)}\frac{dx}{x^4-2x^2}?$$ $\endgroup$ – LutzL May 3 '18 at 11:46
1
$\begingroup$

First, do a change of variable $x = r(z)$. Then do a partial fraction decomposition: $$ \frac{1}{x^4 - 2x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - \sqrt2} +\frac{D}{x + \sqrt2}. $$ Can you continue?

$\endgroup$
  • $\begingroup$ Yes, but I have that $r=r(z)$ and my integral is in form $\int_0^{r(z)} \mathrm{d}z$, what is happening with this dependence $r$ on $z$: $r=r(z)$ in integral with $\mathrm{d}z$? $\endgroup$ – nick_name Mar 1 '18 at 15:08
  • $\begingroup$ If I would use $x=r(z)$ then it would be $\mathrm{d}x=\mathrm{d}r$, but I have only $\mathrm{d}z$ in my equation, what would I do with it on your way? $\endgroup$ – nick_name Mar 1 '18 at 15:40
  • 1
    $\begingroup$ @nick_name, you can calculate the primitive and undo the change of variable or change the limits or integration according to the COV formula: en.wikipedia.org/wiki/…. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 1 '18 at 15:40
  • 1
    $\begingroup$ @nick_name, wrong. $dx = dz$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 1 '18 at 15:41
  • 1
    $\begingroup$ @nick_name, yes. $dx = −(r_i − 1)dz$. I wrote too fast. My point was that the original variable is $z$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 2 '18 at 11:54

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.