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 $u=1+x^2$ :

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

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