Question

What is `int 1/(1+4x^2) dx`?

Answer 1

1/2tan^-1(2x)+C

Explanation:

Start with the trig identity:

`sin^2theta+cos^2theta=1`

If we divide this by `cos^2theta` we get

`sin^2theta/cos^2theta+ cos^2theta/cos^2theta=1/cos^2theta`

`->tan^2theta +1 = sec^2theta`

For integral, we can try the substitution:

`2x = tan(u)` From this we get:
`-> 2dx = sec^2(u) du`

Now putting this substitution into the integral:

`int1/(1+4x^2)dx=1/2int sec^2(u)/(1+tan^2(du))du`

Now using the trig identity on the bottom of the fraction we get:

`1/2intsec^2(u)/sec^2(u)du = 1/2int du`
Integrating then reversing the substitution gives us:

`=1/2u+C=1/2tan^-1(2x)+C`