Answer
$$\int\frac{2^t}{2^t+3}dt=\frac{\ln(2^t+3)}{\ln2}+C$$
Work Step by Step
$$A=\int\frac{2^t}{2^t+3}dt$$
Let $u=2^t+3$
Then we have $du=2^t\ln2dt$. That means $2^tdt=\frac{1}{\ln2}du$
Substitute into $A$: $$A=\frac{1}{\ln2}\int\frac{1}{u}du$$ $$A=\frac{\ln|u|}{\ln2}+C$$ $$A=\frac{\ln|2^t+3|}{\ln 2}+C$$
As all $t\in R$, $2^t\gt0$. That means $(2^t+3)\gt0$.
Therefore, $|2^t+3|=2^t+3$ $$A=\frac{\ln(2^t+3)}{\ln2}+C$$
