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

$\endgroup$
1
  • $\begingroup$ 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
    Commented Sep 14, 2014 at 10:05

1 Answer 1

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: $$ \frac{x^3}{x^4+0.0019x^2+0.002x+0.003}=\frac{a_0 x+b_0}{x^2+\alpha_0x+\beta_0}+\frac{a_1 x+b_1}{x^2+\alpha_1x+\beta_1}. $$

$\endgroup$
2
  • $\begingroup$ thank you , can you please clarify which one is a_0 and a_1 here? $\endgroup$ Commented Sep 14, 2014 at 12:50
  • $\begingroup$ @TazkeraHaqueTrina You may obtain, with the Mathematica command 'Apart[ ]', the decomposition:$$ \frac{-0.0880337+0.514102 x}{0.0579024-0.32835 x+x^2}+\frac{0.078773+0.485898 x}{0.0518113+0.32835 x+x^2}$$ $\endgroup$ Commented Sep 14, 2014 at 13:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .