How do you integrate #(1+ x) / (1 + x^2) dx#?
1 Answer
Jun 9, 2016
Explanation:
We have:
#int(1+x)/(1+x^2)dx#
Split up the numerator and into two different integrals:
#=int1/(1+x^2)dx+intx/(1+x^2)dx#
Notice that the first derivative is just the derivative of the arctangent function, that is,
#=arctan(x)+intx/(1+x^2)dx#
For the remaining integral, let
#=arctan(x)+1/2int(2x)/(1+x^2)dx#
#=arctan(x)+1/2int(du)/u#
This is the natural logarithm integral:
#=arctan(x)+1/2ln(absu)+C#
Since
#=arctan(x)+1/2ln(1+x^2)+C#