Integrating a composite function There was a related question concerning the spectral decomposition of a nested function at [1]. 
Simply put, do the solutions to the integral of a composite function bear any resemblance to the solutions of each function integrated individually? Mine is the same question as the one linked except, without the complex exponential under the integral, it is more general. 
[1] Fourier transform of function composition
 A: No, the antiderivative of a composition need not have any resemblance to the antiderivatives of individual functions in the composition. For example, $e^x$ integrates to $e^x$ and $x^2$ integrates to $\frac13 x^3$, but the integral of $e^{x^2}$ is something entirely different. For a more explicit example, compare $\int \frac{1}{1-x^2}$, $\int \sqrt{x}$, and $\int \sqrt{\frac{1}{1-x^2}}$: the last one is the inverse sine, but the first two have nothing trigonometric in them.
A: While it's true that there is no general formula, one can say something if $g(x)$ happens to be invertible. In this case we can compute the derivative of the inverse by \begin{align*}g(g^{-1}(x))&=x\\g' (g^{-1}(x))\cdot (g^{-1})'(x)&=1\\(g^{-1})'(x)&=\frac{1}{g' (g^{-1}(x))}\end{align*}
Now in the integral we can make the substitution $u=g(x)$ so that $du=g'(x)dx$. Solving in terms of $x$ gives $x=g^{-1}(u)$ and $dx=\frac{du}{g'(x)}=\frac{du}{g'(g^{-1}(u))}=(g^{-1})'(u)du$. Therefore the integral becomes
\begin{align*}\int f(g(x))dx&=\int f(u)(g^{-1})'(u)du\end{align*}
So you could solve this integral and then plug in $u=g(x)$ into the final answer. While this isn't technically in terms of $\int f(x)dx$ and $\int g(x)dx$, it does turn the integrand from being a composition into being a product. Therefore, you could try using integration by parts to get an answer. However, it should be noted that whether or not this formula is actually helpful heavily depends on $f$ and $g$. For example, if $g(x)=e^x$, we get $$\int f(e^x)dx=\int f(u)\ln'(u)du=\int \frac{f(u)}{u}du$$ If $f(x)$ is something simple like $x^n$, then this can be easily computed $$\int \frac{f(u)}{u}du=\int u^{n-1}du=\frac{u^n}{n}=\frac{e^{nx}}{n}$$ On the other hand, in this case the original integral $\int e^{nx}dx$ was easy to begin with. If, as suggested by user53153, we consider $f(x)=e^x$ and $g(x)=x^2$, we get  $$\int e^{x^2}dx=\int \frac{e^u}{2\sqrt{u}}du$$ which is just as difficult to integrate.
