<s>[INST] Calculus Integral $\int \frac{dt}{2^{t} + 4}$ 
I try resolve this integral, with  $u = 2^t + 4 $, but can´t ... plz 
  $$\int \frac{dt}{2^t + 4} = \frac{1}{\ln 2}\int\frac{du}{u(u-4)} = \text{??} $$

 [/INST] $\displaystyle \int\dfrac{dt}{2^t+4}$
on u substitution, i.e., $2^t+4=u,\;$ we get- 
$\dfrac{1}{\ell n2}\displaystyle \int\dfrac{du}{u(u-4)}\;....(1),\;$ as you already done.
Now, we need partial fraction to solve it further.
For this, $\dfrac{1}{u(u-4)}=\dfrac{A}{u}+\dfrac{B}{u-4}$
$=\dfrac{Au-4A+Bu}{u(u-4)}$
$\dfrac{1}{u(u-4)}=\dfrac{(A+B)u-4A}{u(u-4)}$
On comparing L.H.S. and R.H.S.
$(A+B)u=0,\; -4A=1$
$\implies A+B=0,\;A=-\dfrac{1}{4}$
$thus, A=-\dfrac{1}{4}, \;B=\dfrac{1}{4}$
Now, expression (1) becomes, $\dfrac{1}{\ell n2}\displaystyle \int\dfrac{-\dfrac{1}{4}}{u}+\dfrac{\dfrac{1}{4}}{u-4}$
$\dfrac{1}{\ell n2}\displaystyle \int\left(-\dfrac{1}{4u}+\dfrac{1}{4(u-4)}\right)du$
$\dfrac{1}{4\ell n2} \big(\ell n(u-4)-\ell nu\big)$
put the value of u
$\dfrac{1}{4\ell n2} \big(\ell n(2^t)-\ell n(2^t+4)\big)$
$\dfrac{1}{4\ell n2} \big(t\ell n2-\ell n(2^t+4)\big)$
$\dfrac{t}{4}-\dfrac{\ell n(2^t+4)}{4\ell n2}$
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