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:

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

How do you integrate int1/(1+x^2) + int x/(1+x^2) dxint1/(1+x^2) + int x/(1+x^2) dx?