Question

as I can solve this exercise?
2^x/ (4+4^x) dx

Answer 1

integral?

Answer 2

if it an integral
i would go with the substitution u=2^x first

Answer 3

why not u = 2+2^x?

Answer 4

or \[\frac{1}{4 \ln(2)} \int\limits_{}^{}\frac{2^x \ln(2)}{1+(\frac{2^x}{2})^2} dx\]
this will actually combine two sub into one
by doing \[\tan(\theta)=\frac{2^x}{2}\]

Answer 5

If u = 2+2^x, then du = ln2 2^x dx,
du/ln2 = 2^x dx
so that
\[\int \dfrac{1}{2ln2} \dfrac{1}{u}du\]
\(= \dfrac{1}{2ln2}ln(2+2^x)+C\)
Check my stuff, please

Answer 6

what happened to the 4+4^x

Answer 7

oh yea, you are correct, professor. hahaha... learn a lot!! My bad.

Answer 8

\[4+4^x=2 \cdot 2 + (2 \cdot 2)^x\]
oh you tried to factor 2 out

Answer 9

except that one 2 has a x power

Answer 10

yea, I see my mistake.