# Integral of $f(x)/(1+f(x))$

I'm interested in whether or not integrals of the form $I=\int (f(x)/(1+f(x)))\;dx$ exist. In particular, I've been working on polinoms without aby result. Could someone show me how to solve this integral?

$$\int_a^b \frac{f(x)}{1+f(x)} \,dx$$

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The fact that the function to integrate is $g=f/(1+f)$ is no restriction (except that $g(x)\ne1$ for every $x$) as the inversion $f=g/(1-g)$ shows, hence a general result is doubtful. Are you interested in a special class of functions $f$? – Did Jun 17 '12 at 15:17
If $f(x)$ is a polynomial of degree $n$ then $f(x)+1$ will also be. In small cases, one can divide and use $\log$s and $\tan^{-1}$ (maybe up to $\deg 4$), but then things get messy. – Pedro Tamaroff Jun 17 '12 at 15:17

It holds that: $$\frac{f(x)}{1+f(x)}=\frac{1+f(x)-1}{1+f(x)}=1-\frac{1}{1+f(x)}$$

So it also holds that: $$\int\frac{f(x)}{1+f(x)}dx=\int1-\frac{1}{1+f(x)}dx$$

This means we can express the integrand using $f(x)$ just once (instead of twice), which might lead to something we can more easily integrate. For example, take $f(x)=x^2$, then we find that:

$$\int\frac{f(x)}{1+f(x)}dx=\int\frac{x^2}{1+x^2}dx=\int1-\frac{1}{1+x^2}dx=x-\tan^{-1}(x)$$

A more straightforward method to compute this integral would be to use the residue theorem, but that would be more laborious.

As did thoughtfully mentioned, sometimes this trick is counterproductive, such as in the case of $f(x)=e^{cx}$. In this case, the inverse trick would be helpful!

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You have written a true statement which can lead to a meaningful conclusion, but you have not reached one yourself. You should expand your answer. – Aaron Jun 17 '12 at 15:20
But $f(x)=\mathrm e^{cx}$. – Did Jun 17 '12 at 15:35
@did, I don't quite get your comment, could you elaborate on the issue? – akkkk Jun 17 '12 at 15:36
Yes: this is to signal that, for this function, the reduction $f/(1+f)=1-1/(1+f)$ is counterproductive since $f/(1+f)$ has a simple antiderivative while $1/(1+f)$ has not. – Did Jun 17 '12 at 15:38
And in fact, if one form is best for $f$ then the other is best for $1/f$, hence the situation is entirely symmetric. – Did Jun 17 '12 at 15:43