Question

How do you evaluate the integral `int dx/(4+x^2)`?

Answer 1

You may know the following rule:

`intdx/(a^2+x^2)=1/aarctan(x/a)+C`

So:

`intdx/(4+x^2)=intdx/(2^2+x^2)=1/2arctan(x/2)+C`

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We can derive the `intdx/(a^2+x^2)` rule. To do so, let `x=atantheta`.

Thus, `dx=asec^2thetad theta` and `x^2=a^2tan^2theta`.

`intdx/(a^2+x^2)=int(asec^2thetad theta)/(a^2+a^2tan^2theta)`

Factoring the `a^2` terms and using the identity `tan^2theta+1=sec^2theta`:

`=int(asec^2thetad theta)/(a^2(1+tan^2theta))=int(sec^2thetad theta)/(asec^2theta)=1/aintd theta`

The antiderivative of `1` is `theta`, since we're working in terms of `theta` here:

`=1/atheta+C`

From `x=atantheta`, our original substitution, we can solve for `theta`. Note that `tantheta=x/a` so `theta=arctan(x/a)`. Then:

`=1/aarctan(x/a)+C`

Which is the above stated rule.