We have to find the anti derivative of f(x) = 1/(1+x^2)*arc tan x

Here we first take t = arc tan x

=> dt/dx = 1 / ( 1 + x^2)

=> dt = dx / ( 1 + x^2)

Int [ 1/(1+x^2)*arc tan x dx]

=> Int [ (1/t ) dt]

=> ln t + C

replace t with arc tan x

=> ln ( arc tan x ) + C

**Therefore the anti derivative of f(x) = 1/(1+x^2)*arc tan x is ln ( arc tan x ) + C.**

To find the antiderivative F(x), we'll have to evaluate the indefinite integral of f(x).

Int f(x)dx = Int dx/(1+x^2)*arctanx

We'll substitute arctan x = t.

We'll differentiate both sides and we'll get:

dx/(1 + x^2) = dt

We'll re-write the integral:

Int dx/(1+x^2)*arctanx = Int dt/t

Int dt/t = ln|t| + C

We'll substitute t by arctan x:

**Int dx/(1+x^2)*arctanx = ln |arctan x| + C**

To find the anti derivative of f(x)=1/(1+x^2)*arctan x.

Int f(x) dx = Int {dx/(1+x^2)}*arctan x.

Put arctan x = t, dx/(1+x^2 = dt.

Therefore Int f(x) dx = (1/1+x^2)t * dx = tdt.

Int f(x) = t^2/2 = (1/2)(arctan x) +C.

Also if you meant f(x) = 1/{(1+x^2)*arctanx},

Then Int f(x) dx =Int{ 1/[(1+x^2) arctan x]} dx = dt/t.

Int f(x) dx = log t + C.

Int f(x) dx = log (arctan x) + C.