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

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Answer 1 · 2017-01-15T14:18:25

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#

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.