what is integral of this function?

Question

Let $f(x)$ be an arbitrary continuous function, $n\in \mathbb{N}$ and
$$g(x) = \frac{1}{1+n\cdot f(x)^2}$$ then what is anti-derivative of this: $$ \int \left(\frac{d}{dx}g(x)\right)\cdot\tanh\left(n\cdot f(x)\cdot\frac{d}{dx}f(x)\right)dx = ? $$ or $$ \int \left(\frac{d}{dx}g(x)\right)\cdot\tanh\left(n\cdot f(x)\right)\cdot\tanh\left(n\cdot\frac{d}{dx}f(x)\right)dx = ? $$ Both integral equals(in the limit when n goes to infinity). Also we know that $\int \left(\frac{d}{dx}g(x)\right)dx = g(x)$ and $\int \left(n.f(x).\frac{d}{dx}f(x)\right)dx = n\cdot\frac{f(x)^2}{2}$

Answer 1

There is no reason to expect a closed form answer for this problem; take $n=1$ and $f(x)=x^2$, and ask your favorite CAS to do your first antiderivative. Wolfram Alpha states that there is no closed form, and while that is not a proof, it is good enough for me (at least until you can demonstrate that there should be a closed form somehow).