Simplifying $\int \frac{f(x)}{f'(x)}\,\mathrm{d}x$ I have an integral in the form of
$$ \int \frac{f(x)}{f'(x)}\,\mathrm{d}x $$
I am trying to find a way to integrate it. Is there any method or should I keep working on simplifying $f(x)$ and $f'(x)$?
Just in case, $f(x) = x + \sin x$ and, thus, $f'(x) = 1+\cos x$
 A: (In this case) You can rewrite the integral as follows:
$$\int\frac{x+2\sin\frac x2 \cos \frac x2}{2\cos^2\frac x2}\ dx$$
Split the terms to get
$$\int\frac x2\sec^2\frac x2 + \tan\frac x2 \ dx$$
Integrate the first term by parts
$$x\tan\frac x2 - \int\tan \frac x2\ dx + \int \tan \frac x2 \ dx$$
Cancelling the last two terms gives us the result
$$x \tan \frac x2 + c$$
As a general case, there is no way to resolve $\int\frac{f(x)}{f'(x)}dx$ into something that is easily integrable; you would have to resort to simplifying the numerator and evaluating the integral.
A: I did not think much on that problem, but, what if you try this way?
$\int\frac{f(x)}{f'(x)}dx = \int\frac{1}{\frac{f'(x)}{f(x)}}dx = \int\frac{1}{\frac{dln(f(x))}{dx}}dx$
And then, maybe, doing a variable change like this
$\ \ \ ln(f(x)) = y\ \ \ ---> \frac{dln(f(x))}{dx}dx = dy \ \ \ ---> \frac{dx}{\frac{dln(f(x))}{dx}} = \frac{dy}{(\frac{dln(f(x))}{dx})^2}$
you got
$\int\frac{1}{(\frac{dy}{dx})^2}dy = \int(\frac{dx}{dy})^2dy$
I do not know how I can continue. I hope this can be useful for you. Neither I know the domain and image of $f(x)$ for the $ln(f(x))$ be a good function, but, asuming this, it can be a good way.
A: There is no general method to evaluate $\int \frac{f(x)}{f'(x)}dx$, unlike $\int \frac{f'(x)}{f(x)}dx = \ln|f(x)| + c$.
So you have to actually work on the particular integral here.
Multiply the integrand by $\frac{1-\cos x}{1-\cos x}$ and use trigonometric identities like $1 - \cos^2 x = \sin^2 x$ to give:
$\int x \csc^2 x + \csc x - x \frac{\cos x}{\sin^2 x} - \frac{\cos x}{\sin x}dx$, each term of which can be evaluated in an elementary way. Some are textbook integral forms, some require a bit of integration by parts, and the final term is of the form $\int \frac{f'(x)}{f(x)}dx$ (or you can just look up the textbook for the integral of $\cot x$).
A: If you have
$$\int \frac{x+\sin (x)}{1+\cos (x)} dx=\int \frac{\sin (x)}{1+\cos (x)} dx+\int \frac{x}{1+\cos (x)} dx$$
For the first one,
$$\int \frac{\sin (x)}{1+\cos (x)} dx=-\int \frac{d(\cos (x))}{1+\cos (x)} dx=-\log(1+\cos(x))$$
Rewrite the second one
$$\int \frac{x}{1+\cos (x)} dx=={\displaystyle\int}\dfrac{x\csc\left(x\right)}{\csc\left(x\right)+\cot\left(x\right)}\,dx$$ and perform one integration by parts to get
$$\dfrac{x}{\csc\left(x\right)+\cot\left(x\right)}-{\displaystyle\int}\dfrac{dx}{\csc\left(x\right)+\cot\left(x\right)}$$ Now, the tangent half-angle substitution leads to something very simple.
