How do you evaluate the integral of $int x*arctan(x)$ ?

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Answer 1 · 2016-08-12T22:56:46

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

$int x*arctan(x) \ dx$

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

which, by IBP, is.....
$=x^2/2 *arctan(x) - int (x^2/2) d/dx arctan(x) \ dx$

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

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

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

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

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

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

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

How do you evaluate the integral of int x*arctan(x)int x*arctan(x)?