Calculus Integral $\int \frac{dt}{2^{t} + 4}$

Question

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

Answer 1

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.

Answer 2

hint: partial fractions on the $u$ integral

Answer 3

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 $\dfrac{t \ln 2 - \ln (2^t+4)}{4 \ln2}$

Answer 4

Hint: $$\frac{1}{2^x+4}=\frac{2^x-2^x+4}{4(2^x+4)}=\frac{1}{4} \left(1-\frac{2^x}{2^x+4}\right)$$

Answer 5

$\displaystyle \int\dfrac{dt}{2^t+4}$

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

$\dfrac{1}{4\ell n2} \big(\ell n(2^t)-\ell n(2^t+4)\big)$

$\dfrac{1}{4\ell n2} \big(t\ell n2-\ell n(2^t+4)\big)$

$\dfrac{t}{4}-\dfrac{\ell n(2^t+4)}{4\ell n2}$