We have to find Int [1/ (1 + 4x^2) dx].

First substitute u = 2x

=> du /dx = 2

=> du /2 = dx

Now Int [1/ (1 + 4x^2) dx]

=> Int [(1/2)*(1/ (1+u^2) du]

=> (1/2)*Int [1/ (1 + u^2) du]

Now Int [1/ (1+u^2) du] =arc tan u + C

=> (1/2)*arc tan u + C

replace u with 2x

=> (1/2)* arc tan 2x + C

Therefore the required result is

**1/2)* arc tan 2x + C**

To evaluate the integral of 1/(1+4x^2).

Solution:

We know that Integral of 1/(x^2+a^2 ) = Int {1/(x^2+a^2)} dx = (1/a)arc tan (x/a) + C....(1) .

The given function 1/(4x^2+1) = (1/4){1/[x^2+(1/4)]} = (1/4){1/[x^2+(1/2)^2]}.

Therefore Int {1/(4x^2+1)} dx = (1/4) Int {1/[x^2+(1/2)^2]} dx which is like (1). Here a = 1/2.

So Int {1/(4x^2+1)} dx = (1/4)* (1/(1/2))* arc tan {x/(1/2)}+C .

**Therefore Int {1/(4x^2+1)} dx = (1/2) arc tan(2x) +C.**