I'm talking about:
$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{(n+1)(a)}$
$\int\frac{1}{ax+b}dx=\frac{1}{a}\ln(ax+b)$
$\int e^{ax+b}dx=\frac{e^{ax+b}}{a}$
$\int a^{ax+b}dx=\frac{a^{ax+b}}{(\ln|a|)(a)}$
Of course, this only works for $(ax+b)$, not for other $f(x)$ with higher exponents like $x^2$ or $x^3$
So for example if write my working as:
\begin{align} \int(2x+8)^3dx&=\frac{(2x+8)^4}{(4)(2)} \\ &=\frac{1}{8}(2x+8)^4 \end{align}
... based on the rule above
I just want to know if such working/rule applied is actually acceptable without using the substitution method.
Okay I'm going to prove/disprove each one of these bad boys.
#1: $$I=\int(ax+b)^n\mathrm{d}x$$ Substitution: $u=ax+b\Rightarrow \frac{\mathrm{d}u}a=\mathrm{d}x$ $$I=\frac1a\int u^n\mathrm{d}u$$ $$I=\frac{(ax+b)^{n+1}}{a(n+1)}$$ Correct
#2: $$I=\int\frac{\mathrm{d}x}{ax+b}$$ substitution: $u=ax+b\Rightarrow \frac{\mathrm{d}u}a=\mathrm{d}x$ $$I=\frac1a\int\frac{\mathrm{d}u}u$$ $$I=\frac1a\ln|u|$$ $$I=\frac1a\ln|ax+b|, \qquad a\neq0$$ You forgot the $|$ bars, but otherwise correct
#3: $$I=\int e^{ax+b}\mathrm{d}x$$ substitution: $u=ax+b\Rightarrow \frac{\mathrm{d}u}a=\mathrm{d}x$ $$I=\frac1a\int e^u\mathrm{d}u$$ $$I=\frac1ae^u$$ $$I=\frac1ae^{ax+b}$$ Correct
#4: $$I=\int a^{ax+b}\mathrm{d}x$$ $$I=\int e^{(ax+b)\ln a}\mathrm{d}x$$ substitution: $u=(ax+b)\ln a\Rightarrow \frac{\mathrm{d}u}{a\ln a}=\mathrm{d}x$ $$I=\frac1{a\ln a}\int e^u\mathrm{d}u$$ $$I=\frac1{a\ln a}e^u$$ $$I=\frac{e^{(ax+b)\ln a}}{a\ln a}$$ $$I=\frac{a^{ax+b}}{a\ln a}$$ $$I=\frac{a^{ax+b-1}}{\ln a}$$ Basically correct
As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving $$\int\frac{\mathrm{d}x}{x}=\ln|x|$$ every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.