Taking the partial fraction when the variable is exponential

Question

I'm trying to solve the following integral:

$\int\frac{dx}{2^x+3}$

and here's what I've done so far:

Substituting $t=2^x$ we have $dt=2^x\ln 2 dx\implies \frac{dt}{t}=\ln2dx$

Now, I have to take the derivative of the following:

$\frac{1}{\ln 2}\int \frac{dt}{t(t+3)}$

My question is, if I were to take the partial fraction of $\frac{1}{t(t+3)}$, then obviously I would have:

$1\equiv A(t+3)+Bt$

But I'm not sure if I can now substitute $t=-3$ in order to determine $B$, since $t=2^x$ is always positive. But according to Symbolab, that is possible:

Am I missing something?

Answer 1

To find $A$ and $B$ from $1\equiv A(t+3)+Bt$ you need not put $t=-3$. Just compare the coefficients. The coefficient of $t$ on LHS os $A+B$ so we must have $A+B=0$. The constant term gives $3A=1$, so $A=\frac 1 3$ and $B=-\frac 1 3$.

[You can also put any two positive values for $t$ and solve the resulting equations for $A$ and $B$].

Answer 2

I have turned your problem into the problem of computing $$ \int\frac{\mathrm dt}{t(t+3)}. $$ It turns out that, for every real number $t$, you have $1=\frac13(t+3)-\frac 13t$. Therefore, for each $t\in\mathbb{R}\setminus\{0,-3\}$, you have $$ \frac1{t(t+3)}=\frac1{3t}-\frac1{3(t+3)}\tag1 $$
The way that you have proved that $(1)$ holds doesn't matter.