Given f'(x)=ln(1+tanx) what is the function f?

We'll have to determine the primitive of the given
function and we'll do it by integrating f'(x).

The first
step is to replace x by pi/4 - t

x = pi/4 -
t

Differentiating, we'll
have:

dx = -dt

I = Int
ln(1+tanx)dx = -Int ln[1+tan (pi/4 - t)]dt

But tan (pi/4 -
t) = (tan pi/4 - tan t)/(1 + tan (pi/4)*tan t)

tan (pi/4 -
t) = (1 - tan t)/(1+tan t)

We'll add 1 both
sides:

1 + tan (pi/4 - t) = 1 + (1 - tan t)/(1+tan
t)

1 + tan (pi/4 - t) = 2/(1 + tan
t)

We'll take logarithm function both
sides:

ln [1 + tan (pi/4 - t)] = ln [2/(1 + tan
t)]

ln [2/(1 + tan t)] = ln 2 - ln (1 + tan
t)

We'll integrate:

Int [2/(1
+ tan t)]dt = -Int ln 2 dt + Int ln (1 + tan t)dt

I = Int
ln 2 dt - Int ln (1 + tan t)dt

But Int ln (1 + tan t)dt =
I

I = Int ln 2 dt - I

2I =(ln
2)*t

I = [t*(ln 2)]/2 + C

I =
(pi/4 - x)*ln 2/2 + c

I = (pi/8)*ln 2 - x*(ln 2)/2 +
C

The function f(x) is: f(x) = (pi/8)*ln 2 -
x*(ln 2)/2 + C