integrating a function within another function Is there a general solution for the following integration?
$$\int{f(g(x))}{dx}$$
Hypothesis (Probably wrong and you should ignore) from observations:
$$\int{f(g(x))}{dx}=\frac{F(g(x))}{g^{\prime}(x)}+c$$
Where $F(x)=\int{f(x)}{dx}$ and $c$ is a constant
 A: No, there is no general solution for the primitive of $f\circ g$.
The derivative of your proposed solution is 
$$\frac{g'(x) ( g'(x) f(g(x)) ) - F(g(x))g''(x) }{(g'(x))^2} = f(g(x)) - \frac{F(g(x))g''(x) }{(g'(x))^2}$$
If there was a formula with just the primitive of $f$, it would be easy to integrate any function $v$ : choose $f(x) = \ln(x)$ and define $g(x) = e^{v(x)}$. then $f(g(x)) = v(x)$
Another remark : $\exp$ and $-x^2$ are very simple functions to integrate, but the primitive of $e^{-x^2}$ cannot be expressed with a finite number of simple functions 
A: There are no general solution to such an integral. The hypothesis that you proposed is only true for a very small number of cases and perhaps can be refined for those.
Please do not take offense at some of the responses, not everyone have the tendency to understand and forgive those who know less than themselves.
A: That's not always true.
Try taking the derivative of $ \frac{F(g(x))}{g'(x)}$.  Note that you must apply quotient rule.   As an example consider $f(x) = x$ and $g(x) = x^2$ and compute the integral using your formula of $f(g(x)) = x^2$.  
Integrating functions like $f(g(x))$ isn't always easy.   For example, consider $f(x) = e^x$ and $g(x) = -x^2$. The integral of  $f(g(x)) = e^{-x^2}$  isn't an elementary function. 
However, as you might guess, it is easy to integrate functions like $f'(g(x))g'(x)$ since 
$$\int f'(g(x))g'(x) dx = f(g(x) ) +c $$
A: This is incorrect as if you take derivative from both sides of your hypothesis, you get
$$f(g(x))=\frac{f(g(x))\cdot (g'(x))^2-F(g(x))\cdot g''(x)}{(g'(x))^2},$$
Which obviously does not hold.
