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

 A: hint: partial fractions on the $u$ integral
A: With the substitution $u=2^t+4$ we have $du=(\ln 2) 2^t \,dt=\ln2 (u-4)\, dt$
Hence as you correctly mentioned, our integral becomes equivalent to:
$$\frac{1}{\ln 2} \int\frac{du}{u(u-4)}$$
Now the method used to solve this integral, or integrals of this sort, is partial fractions.
$$\frac{1}{u(u-4)}=\frac{A}{u-4}+\frac{B}{u}$$
Multiply by $u(u-4)$ on both sides, ($u=0$ and $u=4$ are not part of our domain) 
$$1=Au+B(u-4)$$
$$\color{red}{0}u+\color{blue}{1}=\color{red}{(A+B)}u\color{blue}{-4B}$$
Equate coefficients.
Hence $\color{blue}{-4B}=\color{blue}{1} \implies B=-1/4$, $\color{red}{A+B}=\color{red}{0} \implies A=1/4$.
So,
$$\frac{1}{u(u-4)}=\frac{1/4}{u-4}-\frac{1/4}{u}$$
From which I believe you can proceed.
A: Hint:
$$\frac{1}{2^x+4}=\frac{2^x-2^x+4}{4(2^x+4)}=\frac{1}{4} \left(1-\frac{2^x}{2^x+4}\right)$$
A: You've done the right move. Then since
$$\frac{1}{u(u+4)}=\frac{1}{4u}-\frac{1}{4(u+4)}$$
You can get the answer $\dfrac{t \ln 2 - \ln (2^t+4)}{4 \ln2}$ 
A: $\displaystyle \int\dfrac{dt}{2^t+4}$
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
$\dfrac{1}{4\ell n2} \big(\ell n(2^t)-\ell n(2^t+4)\big)$
$\dfrac{1}{4\ell n2} \big(t\ell n2-\ell n(2^t+4)\big)$
$\dfrac{t}{4}-\dfrac{\ell n(2^t+4)}{4\ell n2}$
