What is the integral of: 3/(3 + e^x) ?

3 = ( 3 + e^x ) - e^x

...................................................

Hence

I = INT { [ ( 3 + e^x ) - e^x ] / ( 3 + e^x ) } dx

= INT { 1 -[ e^x / ( 3 + e^x ) ] } dx

= x - INT [ f ' (x) / f(x) ] dx,..................where f(x) = 3 + e^x

= x - ln [ f(x) ] + C

= x - ln ( 3 + e^x ) + C

Substitute u = 3 + e^x, so that du = e^x dx.

the integral, 3/(3+e^x) times (e^x/e^x)dx, becomes:

3/u times du/(u-3)

The integrand, 3/u(u-3) is integratable by the method of partial fractions; the rational function is equal to -1/u + 1/(u-3), which integrates to become -ln u + ln(u-3) + C

This is equivalent to -ln(3+e^x) + x + C

Fib

3 / (3 + e^x)

= (3 + e^x - e^x) / (3 + e^x)

= (3 + e^x) / (3 + e^x) - e^x / (3 + e^x)

= 1 - e^x / (3 + e^x)

∫ 3 / (3 + e^x)dx

= ∫ (1 - e^x / (3 + e^x)) dx

= x - ln(3+e^x) + C

int 3dx / (3 + e^x) = int 3e^{-x}dx / (3e^{-x} + 1)

= -ln(3e^{-x} + 1) + C

&&&&&&&&&&&&&&&&&&&&&&&

This is equivalent to certain answers above because

-ln(3e^{-x} + 1) + C = -ln[(3 + e^x) / e^x] + C

= -ln(3 + e^x) + ln e^x + C

= x - ln(3 + e^x) + C

multiply by e^x/e^x and then make a substitution u=e^x