Steps for solving a simple quotient integral containing a product in the denominator. lets say that I have these two integrals:
$\int \frac{1}{e^{2x}-2e^x-3} \, dx$ 
and
$\int \frac{1}{(x+1)(x+2)(x-3)} \, dx$ 
I do recognize some properties and antiderivatives involved but wasn't successful by applying $u$-Substitution and I don't know if/how to integrate by parts with more than two functions involved (second example).
What is a tipical approach in these cases?
 A: Based on what you've seen in algebra before calculus, the expression $e^{2x}+2e^x-3$ should make you think of $u=e^x$ so that $e^{2x}+2e^x-3$ becomes $u^2+2u-3 = (u+3)(u-1)$.  But next you need to say $du = e^x\,dx = u\,dx$ so $du/u = dx$.  Then you have
$$
\int \frac{du}{u(u+3)(u-1)} = \int \left( \frac A u + \frac B {u+3} + \frac C {u-1} \right) du
$$
and you need to find $A$, $B$, and $C$.  The second integral you mention can be done similarly by partial fractions.
A: For the first integral write it as follows:$$\int  \frac { dx }{ e^{ 2x }-2e^{ x }-3 } \, =\int { \frac { dx }{ \left( { e }^{ x }-3 \right) \left( { e }^{ x }+1 \right)  }  } =\frac { 1 }{ 2 } \left[ \int { \frac { dx }{ { e }^{ x }-3 } -\int { \frac { dx }{ { e }^{ x }+1 }  }  }  \right] $$
then substitute:${ e }^{ x }=t\Rightarrow \quad x=\ln { t\quad \quad \Rightarrow  } dx=\frac { dt }{ t } $
in order to write as:
$$\frac { 1 }{ 2 } \left[ \int { \frac { dt }{ t\left( t-3 \right)  }  } -\int { \frac { dt }{ t\left( t+1 \right)  }  }  \right] $$ 
