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

1 Answer
Jun 26, 2015

#arctan(x)+C#

Explanation:

This is one to memorize:

#\int 1/(1+x^2)\ dx=arctan(x)+C#

It can be derived by differentiation of both sides of the equation #tan(arctan(x))=x# (assuming you know that #d/dx(tan(x))=sec^{2}(x)#) and using the Chain Rule:

#sec^{2}(arctan(x))*d/dx(arctan(x))=1#

#\Rightarrow d/dx(arctan(x))=1/sec^[2}(arctan(x))#

#=1/(1+tan^{2}(arctan(x)))=1/(1+x^2)# for all #x\in RR#.