How do you integrate #(1+ x) / (1 + x^2) dx#?

1 Answer
Jun 9, 2016

#arctan(x)+1/2ln(1+x^2)+C#

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, #int1/(1+x^2)dx=arctan(x)+C#.

#=arctan(x)+intx/(1+x^2)dx#

For the remaining integral, let #u=1+x^2# and #du=2xdx#.

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

#=arctan(x)+1/2int(du)/u#

This is the natural logarithm integral: #int(du)/u=ln(absu)+C#

#=arctan(x)+1/2ln(absu)+C#

Since #u=1+x^2#:

#=arctan(x)+1/2ln(1+x^2)+C#