Question

Question #a49f2

Answer 1

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

Explanation:

`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`