find ∫ ((arctan(x))/x)dx


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Question Number 86955 by abdomathmax last updated on 01/Apr/20

$f i n d \int \frac{a r c t a n (x)}{x} d x$
Commented by Ar Brandon last updated on 01/Apr/20

$L e t f (a) = \int \frac{a r c t a n (a x)}{x} d x$ $\Rightarrow f^{'} (a) = \int \frac{x}{x} \cdot \frac{1}{1 + {(a x)}^{2}} d x = \int \frac{1}{1 + {(a x)}^{2}} d x$ $\Rightarrow f^{'} (a) = \frac{1}{a} \cdot a r c t a n (a x) + C$ $. . . . .$
	Answered by redmiiuser last updated on 01/Apr/20

	$a r c \tan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + . . .$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . x^{2 n + 1}}{2 n + 1}$ $\int \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . x^{2 n} . d x}{2 n + 1}$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . x^{2 n + 1}}{{(2 n + 1)}^{2}}$
	Commented by redmiiuser last updated on 01/Apr/20

	$M r . a b d o m a t h m a x p l s$ $c h e c k m y a n s w e r a n d$ $c o m m e n t t h e a p p r o p r i a t e .$
	Commented by mathmax by abdo last updated on 01/Apr/20

	$y o u r a n s w e r i s c o r r e c t b e c a u s e t h e r a d i u s o f a r c t a n (x) i s + \infty$ $(i n t e g r s e r i e)$
	Answered by mind is power last updated on 02/Apr/20
	$=ln(x)arctan(x)−∫((ln(x))/(1+x^2 ))dx =ln(x)arctan(x)−∫((ln(x)dx)/((x+i)(x−i))) =ln(x)arctan(x)−(1/(2i))∫{((ln(x))/(x−i))−((ln(x))/(x+i))}dx =arctan(x)ln(x)−(1/(2i))∫((ln(x)dx)/(x−i))+(1/(2i))∫((ln(x))/(x+i))dx ∫((ln(x))/(x+a))dx=∫((ln(y−a))/y)dy=∫((ln(−a)+ln(1−(y/a)))/y)dy =ln(−a)ln(x+a)−Li_2 ((x/a)+1) =arctan(x)ln(x)−(1/(2i))ln(i)ln(x−i)+(1/(2i))Li_2 (ix+1)+(1/(2i))ln(−i)ln(x+i) −(1/(2i))Li_2 (−ix+1)+c$
	$= l n (x) a r c t a n (x) - \int \frac{l n (x)}{1 + x^{2}} d x$ $= l n (x) a r c t a n (x) - \int \frac{l n (x) d x}{(x + i) (x - i)}$ $= l n (x) a r c t a n (x) - \frac{1}{2 i} \int {\frac{l n (x)}{x - i} - \frac{l n (x)}{x + i}} d x$ $= a r c t a n (x) l n (x) - \frac{1}{2 i} \int \frac{l n (x) d x}{x - i} + \frac{1}{2 i} \int \frac{l n (x)}{x + i} d x$ $\int \frac{l n (x)}{x + a} d x = \int \frac{l n (y - a)}{y} d y = \int \frac{l n (- a) + l n (1 - \frac{y}{a})}{y} d y$ $= l n (- a) l n (x + a) - {L i}_{2} (\frac{x}{a} + 1)$ $= a r c t a n (x) l n (x) - \frac{1}{2 i} l n (i) l n (x - i) + \frac{1}{2 i} {L i}_{2} (i x + 1) + \frac{1}{2 i} l n (- i) l n (x + i)$ $- \frac{1}{2 i} {L i}_{2} (- i x + 1) + c$
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