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

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Answer 1 · 2016-06-09T23:02:53

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

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#