I have to integrate a specific function using substitution. The problem refers to Schaum's Mathematical handbook and to a formula 16.6. I understand how to substitute a variable but I do not understand the last part of the 16.6 formula where $\displaystyle\int \frac{F(u)}{f'(x)}\,\mathrm du$. I have substituted $f(x)$ with $t$ and $\mathrm dx$ with $g(t)\,\mathrm dt$ but dividing $F(t)$ with $f'(x)$ is what troubles me.
---UPDATE---
the formula 16.6 is $\int F(f(x))dx =\int F(u)\frac{dx}{du}du =\int \frac{F(u)}{f'(x)}du$, where u=f(x). I have a function $x^2cos(ln(4x+3)^{1/4})$ which I need to integrate. I've substituted $ln(4x+3)^{1/4}$ with t, x with $\frac{e^{4t}-3}{4}$ and dx with $e^{4t}$dt. What I don't understand is how to implement the formula 16.6 as I don't get the last part of it.
If you have $u = f(x)$, then the differential is $du = f'(x)\; dx.$ If you solve that for $dx$ you get
$$dx = \frac{du}{f'(x)).$$
If you substitute this for $dx$ in the second integral in Formula 16.6, you get the third integral.
I can't tell whether the $1/4$ power in supposed to be inside or outside the $\ln$. If it's outside, then Maple won't do the integral. If it's inside, then Maple will do it, but the answer is obscene.