Question

integrate 2^x/2^x+1

Answer 1

let u=2^x+1
du=ln2*2^x dx
1/ln2 du=2^x dx

Answer 2

\[\frac{1}{\ln2}\int\limits_{}^{}\frac{du}{u}\]

Answer 3

\[\frac{1}{\ln2}*lnu+C\]

Answer 4

\[\frac{1}{\ln2}\ln{(2^x+1)}+C\]

Answer 5

hey 2^x+1 is in the denominator right?

Answer 6

\[\int\limits_{}^{}2^x/(2^x+1)dx\]Let \[u=2^x\]Then \[du=2^x \ln 2dx\]Or,\[du/\ln2=2^xdx\]Plugging this in to the integral we have\[1/\ln2\int\limits\limits_{}^{}du/(u+1)\]Thus,\[=\ln (u+1)/\ln2 + C\] or \[\ln(2^2+1)/\ln2 +C\]

Answer 7

i think eseild meant for that exponent inside the natural log to be a x not a 2

Answer 8

yeah oooops

Answer 9

typo lol

Answer 10

so can i leave the answer like 1/ln2*lnI2^x+1I+c

Answer 11

Yeah anil, there isn't anything you can do to simplify it honestly.