MHB Do you say this integration is doable?

[clumps tim]

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

 

[I like Serena]

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

Hi cooper607! Welcome to MHB! :)

I'm afraid your integral is undefined.
If we pick $\beta=0$, we would get:
$$\nu = \int_0^{P(r)} \frac{dP}{P} = \ln({P(r)}) - \ln(0) = \infty$$
Perhaps the lower boundary should be for instance $P_0$?

Furthermore, I do not think your integral can be solved symbolically.

However, seeing that it appears to be an engineering approximation, it seems to me that is probably not really necessary.

What we can do is make for instance the following approximation:
\begin{aligned}\nu &=\int_{P_0}^{P(r)} \frac{dP}{P+\beta\rho(P)} \\
&= \int_{P_0}^{P(r)} \frac{1}{P} \cdot \frac{dP}{1+\frac{\beta\rho(P)}{P}} \\
&\approx \int_{P_0}^{P(r)} \frac{1}{P}\left(1-\frac{\beta\rho(P)}{P}\right)\, dP \\
&= \int_{P_0}^{P(r)} \left(\frac{1}{P}-\frac{\beta\rho(P)}{P^2}\right)\, dP \\
\end{aligned}
From here on integration is straight forward.
The result is of the form:
$$\nu = \ln\left(\frac{P(r)}{P_0}\right) + \text{small correction}$$