## Integral Equation

Any advice on how to approach a problem like this either numerically or analytically? I'm looking to find the form of f where b>a.

$$\frac{1}{f(r)+1} = \int_r^b f(x) dx + \int_a^r \frac{x^2}{r^2}f(x)dx$$

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 Hey DukeLuke. The first suggestion would be to differentiate both sides and use the fundamental theorem of calculus to get everything in terms of f(x) and the other terms. Once you get the differential equation for f and r, you can then at a minimum use numerical techniques to obtain an approximation, For the LHS you can use standard quotient rules and collect terms and for integrals you use the fundamental theorem of calculus. Have you come across the FTC for integral expressions and relationship between derivatives?
 Thanks for the response. I'm confused about how to apply the FTC to the second integral. I'm not sure what $$\frac{d}{dr} \int_a^r \frac{x^2}{r^2}f(x) dx$$ is because it contains the $x^2$ and $r^2$ in the integrand. It seems like some sort of product rule for integration is needed (or integration by parts without knowing the form of the function). I think the other terms are $$\frac{d}{dr} \int_r^b f(x) dx = -f(r)$$ $$\frac{d}{dr} \frac{1}{f(r)+1} = \frac{-f^'(r)}{(f(r)+1)^2}$$

## Integral Equation

 Quote by DukeLuke Thanks for the response. I'm confused about how to apply the FTC to the second integral. I'm not sure what $$\frac{d}{dr} \int_a^r \frac{x^2}{r^2}f(x) dx$$
You need to apply Leibnitz's rule to that:

$$\frac{d}{dr} \int_a^r \frac{x^2}{r^2}f(x) dx =f(r)+\int_a^r \frac{\partial}{\partial r} \left(\frac{x^2}{r^2} f(x)\right)dx$$

I don't think that's going to do it though guys assuming $a\leq r\leq b$.

 If I apply the Liebniz rule to the term I'm having trouble with, $$\frac{d}{dr} \int_a^r \frac{x^2}{r^2} f(x) dx = f(r) - 2 \int_a^r \frac{x^2}{r^3} f(x) dx$$ The f(x) cancels with the other integral result of -f(x). If I differentiate one more time using the rule I think I'm left with, $$\frac{d}{dr} \int_a^r \frac{ - 2x^2}{r^3} f(x) dx = -2\frac{f(r)}{r} +6 \int_a^r \frac{x^2}{r^4} f(x) dx$$ In my case it's safe to assume that x and r are of the same order meaning that $\frac{x^2}{r^4}$ will be small. Do you think I can safely approximate the differential equation with $$\frac{d}{dr} \frac{-f^'(r)}{(f(r)+1)^2} = -2\frac{f(r)}{r}$$ Edit: just saw jackmell's post and yes a

 Quote by DukeLuke Do you think I can safely approximate the differential equation with $$\frac{d}{dr} \frac{-f^'(r)}{(f(r)+1)^2} = -2\frac{f(r)}{r}$$ Edit: just saw jackmell's post and yes a
I'm all about trying. I would try that, obtain a solution, any reasonable solution, then back-substitute it into the original integral equation and see how the left and right sides agree. If they agree reasonably well for my particular application, I would be satisfied.

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 Quote by DukeLuke If I apply the Liebniz rule to the term I'm having trouble with, $$\frac{d}{dr} \int_a^r \frac{x^2}{r^2} f(x) dx = f(r) - 2 \int_a^r \frac{x^2}{r^3} f(x) dx$$ The f(x) cancels with the other integral result of -f(x). If I differentiate one more time using the rule...
$$\frac{1}{f(r)+1} = - 2 \int_a^r \frac{x^2}{r^3} f(x) dx$$
$$\frac{r^3}{f(r)+1} = - 2\int_a^r x^2 f(x) dx$$