Question

\[\huge \int \frac{2^t}{2^t + 3} dt\]

Answer 1

Add +3 and -3 to the numerator..

Answer 2

hmm?

Answer 3

Then separate them and you will have your answer..

Answer 4

you wanna do \[\huge \int dt - 3\int frac{1}{2^t + 3}dt?\]

Answer 5

Yes I think..

Answer 6

\[\huge \int dt - 3\int \frac{1}{2^t + 3}dt\]
lol

Answer 7

It's a u substitution:
\[u=2^t +3 \implies du = \ln(2)2^t dt \implies \frac{du}{\ln(2)}=2^tdt\]
\[\implies \frac{1}{\ln2}\int\limits \frac{du}{u} = \frac{1}{\ln2}\ln|2^t+3|+C\]

Answer 8

\(\huge\frac{1}{2^t + 3} \) can be written as \((2^t + 3)^{-1}\)

Answer 9

ahh good one! @malevolence19 thanks

Answer 10

why didnt i think of that lol

Answer 11

Or you can go with malevolence method..