2
$\begingroup$

Hi I have an integral to do

$$\nu =\int_{0}^{P(r)} \,\frac{dP}{P+\beta\rho(P)}$$

here I calculated

$$\rho = 0.003 P^{\frac{2}{4}}+ 0.002P^{\frac{2.5}{4}}+0.0019P^{\frac{3}{4}}$$

My question can this integral be solved anyhow? I tried it in wolfram but it failed, can anyone give me the command in mathematica 10 to solve the integral part only ? I will later put the limits.

reagrds

$\endgroup$
  • $\begingroup$ 9 times it of 10 if wolfram can not find it neither can the populous. That being said I would be happy to see a solution analytically.. $\endgroup$ – Chinny84 Sep 14 '14 at 10:05
1
$\begingroup$

Hint: By the change of variable $$P=x^8, \qquad dP=8x^7dx$$ you readily obtain $$ \begin{align} \nu &=\int_{0}^{P(r)} \,\frac{dP}{P+\beta\rho(P)}\\\\ &=8 \int_{0}^{\sqrt[8]{P(r)}} \,\frac{x^7}{x^8+\beta \left(0.0019x^6+0.002x^5+0.003x^4\right)}{\rm d}x\\\\ &=8 \int_{0}^{\sqrt[8]{P(r)}} \,\frac{x^3}{x^4+0.0019x^2+0.002x+0.003}{\rm d}x\\ \end{align} $$ and then conclude by a partial fraction decomposition of the form: $$ \frac{x^3}{x^4+0.0019x^2+0.002x+0.003}=\frac{a_0 x+b_0}{x^2+\alpha_0x+\beta_0}+\frac{a_1 x+b_1}{x^2+\alpha_1x+\beta_1}. $$

$\endgroup$
  • $\begingroup$ thank you , can you please clarify which one is a_0 and a_1 here? $\endgroup$ – Tazkera Haque Trina Sep 14 '14 at 12:50
  • $\begingroup$ @TazkeraHaqueTrina You may obtain, with the Mathematica command 'Apart[ ]', the decomposition:$$ \frac{-0.0880337+0.514102 x}{0.0579024-0.32835 x+x^2}+\frac{0.078773+0.485898 x}{0.0518113+0.32835 x+x^2}$$ $\endgroup$ – Olivier Oloa Sep 14 '14 at 13:10

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.