How do you integrate #int1/(1+x^2) + int x/(1+x^2) dx#?

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Answer 1 · 2018-04-01T17:29:25

#int dx/(1+x^2) + int x/(1+x^2) = arctanx + ln sqrt (1+x^2) +C#

The first integral is a well known anti-derivative:

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

For the second we can substitute:

#x^2 = u#

#2xdx = du#

so:

#int x/(1+x^2) = 1/2 int (du)/(1+u) = ln sqrt abs (1+u)+C#

then:

#int dx/(1+x^2) + int x/(1+x^2) = arctanx + ln sqrt (1+x^2) +C#