∫ x(arctan x)^2 dx=?


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Question Number 155897 by cortano last updated on 05/Oct/21

$\int x {(\arctan x)}^{2} dx = ?$
	Answered by puissant last updated on 05/Oct/21
	$Q=∫x(arctanx)^2 dx.. IBP ⇒ .. { ((u=arctan x)),((v′=xarctan x)) :} ⇒ { ((u′=(1/(1+x^2 )))),((v=−(x/2)+(1/2)(1+x^2 )arctanx)) :} ⇒ Q=−(x/2)arctanx+(1/2)(1+x^2 )arctan^2 x−(1/2)∫{arctanx+(x/(1+x^2 ))}dx =−(x/2)arctanx+(1/2)(1+x^2 )arctan^2 x−(1/2)(xarctanx−(1/2)ln(1+x^2 ))+(1/4)ln(1+x^2 )+C ∴∵ Q=(1/2)(1+x^2 )arctanx−x arctanx+(1/2)ln(1+x^2 )+C... ...........................Le puissant........................$
	$Q = \int x {(a r c t a n x)}^{2} d x . .$ $I B P \Rightarrow . .$ ${\begin{cases} u = a r c t a n x \\ v^{'} = x a r c t a n x \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{1}{1 + x^{2}} \\ v = - \frac{x}{2} + \frac{1}{2} (1 + x^{2}) a r c t a n x \end{cases}$ $\Rightarrow Q = - \frac{x}{2} a r c t a n x + \frac{1}{2} (1 + x^{2}) {a r c t a n}^{2} x - \frac{1}{2} \int {a r c t a n x + \frac{x}{1 + x^{2}}} d x$ $= - \frac{x}{2} a r c t a n x + \frac{1}{2} (1 + x^{2}) {a r c t a n}^{2} x - \frac{1}{2} (x a r c t a n x - \frac{1}{2} l n (1 + x^{2})) + \frac{1}{4} l n (1 + x^{2}) + C$ $∴∵ Q = \frac{1}{2} (1 + x^{2}) a r c t a n x - x a r c t a n x + \frac{1}{2} l n (1 + x^{2}) + C . . .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . . . . . . . . . .$
	Commented by SANOGO last updated on 06/Oct/21

	$l e d u r g a r$
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