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Answer 1 · 2016-09-24T21:30:23

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

$I = int x arctan(x^2) dx$

we first note that: $arctan u = int 1/(1+u^2) du$

Or IOW, $d/(du) arctan u = 1/(1+u^2)$ so it's a manageable function

we can then say, setting up an IBP, that:

$I int x arctan(x^2) dx$

$=int d/dx(x^2/2) arctan(x^2) dx$

$=x^2/2 arctan(x^2) -int x^2/2 d/dx(arctan(x^2)) dx$

$implies I =x^2/2 arctan(x^2) - color(red)(int x^2/2 1/(1+x^2) dx) qquad triangle$

if we take the red term, we have:

$1/2 int (x^2)/(1+x^2) dx$

$= 1/2 int (1 + x^2 - 1)/(1+x^2) dx$

$= 1/2 int 1 - ( 1)/(1+x^2) dx$

$=color(red)( 1/2 ( x - arctan (x) )+ C)$ which we can plug back into $triangle$ to give:

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

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