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

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!
