∫_0 ^1 ((tan^(−1) x)/(1+x))dx


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Question Number 70718 by oyemi kemewari last updated on 07/Oct/19

$\int_{0}^{1} \frac{\tan^{- 1} x}{1 + x} dx$
Commented by mathmax by abdo last updated on 07/Oct/19
$let I =∫_0 ^1 ((arctan(x))/(1+x))dx and f(t)=∫_0 ^1 ((arctan(tx))/(1+x))dx with t≥0 I=f(1) and f^′ (t)= ∫_0 ^1 (x/((1+x^2 t^2 )(1+x)))dx =_(xt=u) ∫_0 ^t (u/(t(1+u^2 )(1+(u/t)))) (du/t) =∫_0 ^t (u/(t(1+u^2 )(t+u)))du =(1/t) ∫_0 ^t ((udu)/((u+t)(u^2 +1))) let decompose F(u)=(u/((u+t)(u^2 +1))) F(u)=(a/(u+t)) +((bu +c)/(u^2 +1)) a=lim_(u→−t) (u+t)F(u)=((−t)/(t^2 +1)) lim_(u→+∞) uF(u)=0 =a+b ⇒b=(t/(t^2 +1)) ⇒ F(u)=((−t)/((t^2 +1)(u+t))) +((((tu)/(t^2 +1))+c)/(u^2 +1)) F(0)=0=((−t)/(t(t^2 +1))) +c =−(1/(t^2 +1)) +c ⇒c=(1/(t^2 +1)) ⇒ F(u)=(1/(t^2 +1)){((−t)/(u+t)) +((tu +1)/(u^2 +1))} ⇒ f^′ (t)=∫_0 ^t F(u)du =((−t)/(t^2 +1)) ∫_0 ^t (du/(u+t)) +(1/(t^2 +1))(t/2) ∫_0 ^t ((2u)/(u^2 +1))du+(1/(t^2 +1))∫_0 ^t (du/(u^2 +1)) =−(t/(t^2 +1))[ln(u+t)]_0 ^t +(t/(2(t^2 +1)))[ln(u^2 +1)]_0 ^t +((arctan(t))/(t^2 +1)) =((−tln(2))/(t^2 +1)) +((tln(t^2 +1))/(2(t^2 +1))) +((arctan(t))/(t^2 +1)) ⇒ f(t)=−ln(2)∫_0 ^t (x/(x^2 +1))dx +∫_0 ^t ((xln(x^2 +1))/(2(x^2 +1)))dx +∫_0 ^t ((arctan(x))/(x^2 +1))dx +c .....be continued...$
$l e t I = \int_{0}^{1} \frac{a r c t a n (x)}{1 + x} d x a n d f (t) = \int_{0}^{1} \frac{a r c t a n (t x)}{1 + x} d x w i t h t ⩾ 0$ $I = f (1) a n d f^{^{'}} (t) = \int_{0}^{1} \frac{x}{(1 + x^{2} t^{2}) (1 + x)} d x$ $=_{x t = u} \int_{0}^{t} \frac{u}{t (1 + u^{2}) (1 + \frac{u}{t})} \frac{d u}{t} = \int_{0}^{t} \frac{u}{t (1 + u^{2}) (t + u)} d u$ $= \frac{1}{t} \int_{0}^{t} \frac{u d u}{(u + t) (u^{2} + 1)} l e t d e c o m p o s e F (u) = \frac{u}{(u + t) (u^{2} + 1)}$ $F (u) = \frac{a}{u + t} + \frac{b u + c}{u^{2} + 1}$ $a = {l i m}_{u \to - t} (u + t) F (u) = \frac{- t}{t^{2} + 1}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b \Rightarrow b = \frac{t}{t^{2} + 1} \Rightarrow$ $F (u) = \frac{- t}{(t^{2} + 1) (u + t)} + \frac{\frac{t u}{t^{2} + 1} + c}{u^{2} + 1}$ $F (0) = 0 = \frac{- t}{t (t^{2} + 1)} + c = - \frac{1}{t^{2} + 1} + c \Rightarrow c = \frac{1}{t^{2} + 1} \Rightarrow$ $F (u) = \frac{1}{t^{2} + 1} {\frac{- t}{u + t} + \frac{t u + 1}{u^{2} + 1}} \Rightarrow$ $f^{^{'}} (t) = \int_{0}^{t} F (u) d u = \frac{- t}{t^{2} + 1} \int_{0}^{t} \frac{d u}{u + t} + \frac{1}{t^{2} + 1} \frac{t}{2} \int_{0}^{t} \frac{2 u}{u^{2} + 1} d u + \frac{1}{t^{2} + 1} \int_{0}^{t} \frac{d u}{u^{2} + 1}$ $= - \frac{t}{t^{2} + 1} {[l n (u + t)]}_{0}^{t} + \frac{t}{2 (t^{2} + 1)} {[l n (u^{2} + 1)]}_{0}^{t} + \frac{a r c t a n (t)}{t^{2} + 1}$ $= \frac{- t l n (2)}{t^{2} + 1} + \frac{t l n (t^{2} + 1)}{2 (t^{2} + 1)} + \frac{a r c t a n (t)}{t^{2} + 1} \Rightarrow$ $f (t) = - l n (2) \int_{0}^{t} \frac{x}{x^{2} + 1} d x + \int_{0}^{t} \frac{x l n (x^{2} + 1)}{2 (x^{2} + 1)} d x + \int_{0}^{t} \frac{a r c t a n (x)}{x^{2} + 1} d x$ $+ c . . . . . b e c o n t i n u e d . . .$
	Answered by mind is power last updated on 07/Oct/19

	$x = tg (u)$ $\Rightarrow d x = \frac{du}{\cos^{2} (u)}$ $I = \int_{0}^{1} \frac{\tan^{- 1} (x)}{1 + x} dd = \int_{0}^{\frac{π}{4}} \frac{u}{(1 + t g (u))} . \frac{1}{{c o s}^{2} (u)} = \int_{0}^{\frac{π}{4}} \frac{u d u}{(s i n (u) + c o s (u)) (c o s (u))}$ $c o s (u) + s i n (u) = \sqrt{2} (c o s (u - \frac{π}{4}))$ $I = \int_{0}^{\frac{π}{4}} \frac{u d u}{\sqrt{2} c o s (u - \frac{π}{4}) c o s (u)}$ $l e t f (u) = \frac{u}{\sqrt{2} c o s (u - \frac{π}{4}) c o s (u)}$ $i w i l l u s e t h i s i n t h e n e x t s t e p \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ $\Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\frac{π}{4} - u}{\sqrt{2} c o s (\frac{π}{4} - u - \frac{π}{4}) . c o s (\frac{π}{4} - u)} d u$ $I = \int_{0}^{u} \frac{- u d u}{\sqrt{2} c o s (u) c o s (\frac{π}{4} - u)} + \frac{π}{4} \int_{0}^{\frac{π}{4}} \frac{d u}{\sqrt{2} c o s (\frac{π}{4} - u) c o s (u)}$ $\Rightarrow I = - \frac{1}{\sqrt{2}} . I + \frac{π}{4} \int_{0}^{\frac{π}{4}} \frac{d u}{(c o s (u) + s i n (u)) c o s (u)}$ $\Rightarrow I = \frac{\sqrt{2}}{\sqrt{2 +} 1} . \frac{π}{4} \int_{0}^{\frac{π}{4}} \frac{1 . d u}{(c o s (u) + s i n (u)) c o s (u)}$ $1 = {(c o s (u) + s i n (u))}^{2} - 2 s i n (u) c o s (u)$ $A = \int_{0}^{\frac{π}{4}} \frac{1 . d u}{(c o s (u) + s i n (u)) c o s (u)} = \int_{0}^{\frac{π}{4}} \frac{{(c o s (u) + s i n (u))}^{2} - 2 c o s (u) s i n (u)}{(s i n (u) + c o s (u)) c o s (u)}$ $= \int_{0}^{\frac{π}{4}} \frac{c o s (u) + s i n (u)}{c o s (u)} d u - \int_{0}^{\frac{π}{4}} \frac{2 s i n (u)}{c o s (u) + s i n (u)} d u$ $s i n (u) = - (c o s (u) - s i n (u) + (s i n (u) + c o s (u))$ $= \int_{0}^{\frac{π}{4}} (1 + t g (u)) d u - \int_{0}^{\frac{π}{4}} \frac{- (c o s (u) - s i n (u)) + (s i n (u) + c o s (u)) d u}{c o s (u) + s i n (u)}$ $\int_{0}^{\frac{π}{4}} (1 + t g (u)) d u + \int_{0}^{\frac{π}{4}} \frac{c o s (u) - s i n (u)}{c o s (u) + s i n (u)} d u - \int 1 d u$ $= \int_{0}^{\frac{π}{4}} t g (u) d u + \int_{0}^{\frac{π}{4}} \frac{d (s i n (u) + c o s (u))}{c o s (u) + s i n (u)}$ $=_{0}^{\frac{π}{4}} [- l n (c o s (u)) + l n (c o s (u) + s i n (u))]$ $= - l n (\frac{\sqrt{2}}{2}) + l n (\sqrt{2}) = l n (2)$ $I = \frac{\sqrt{2}}{\sqrt{2} + 1} . \frac{π}{4} . A = \frac{π . l n (2) \sqrt{2}}{4 (\sqrt{2} + 1)}$
	Commented by oyemi kemewari last updated on 07/Oct/19
	thank you sir
	Commented by mind is power last updated on 07/Oct/19

	$y^{'} r e w e l c o m$
	Commented by oyemi kemewari last updated on 08/Oct/19
	thank you sir
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