Question

What is the antiderivative of `(1)/(1+x^2)`?

Answer 1

The antiderivative of `1/(1+x^2)` is the integral

`int 1/(1+x^2)dx` which is equivalent to

`int 1/(1+x^2)dx=arctanx+C`

where `arctanx` is the inverse of the trigonometric function

`tanx` and C is the integration constant.

Answer 2

`= arctan(x) + c`

Explanation:

Let

`color(blue)(x = tantheta`

`=> color(red)(dx = sec^2 theta d theta) " By the use of the quotient rule..."`

`int 1/(1+color(blue)(x)^2 ) color(red)(dx) = int 1/(1+color(blue)((tantheta))^2 ) * color(red)(sec^2 theta d theta`

We know `sin^2 x + cos^2 x -= 1`

`=> sin^2 x / cos^2 x + cos^2x/cos^2x -= 1/cos^2 x`

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

`=> int sec^2 theta / sec^2 theta d theta`

`=> int 1 d theta`

`=> theta + c`

If `x = tan theta => arctanx = theta`

Substitute back in...

`color(blue)( arctan(x ) + c`