∫_( 2) ^( 3) ((tan^(−1) (x))/(1 − x^2 )) dx


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Question Number 72665 by TawaTawa last updated on 31/Oct/19

$\int_{2}^{3} \frac{\tan^{- 1} (x)}{1 - x^{2}} dx$
Commented by mind is power last updated on 31/Oct/19

$non close value we have to use {Li}_{2} (x) function$
Commented by TawaTawa last updated on 31/Oct/19

$Help sir$
Commented by mathmax by abdo last updated on 31/Oct/19
$let I =∫_2 ^3 ((arctan(x))/(1−x^2 ))dx =_(x=2+t) ∫_0 ^1 ((arctan(2+t))/(1−(2+t)^2 ))dt =−∫_0 ^1 ((arctan(2+t))/((t+2)^2 −1))dt =−∫_0 ^1 ((arctan(t+2))/((t+1)(t+3)))dt =−(1/2)∫_0 ^1 arctan(t+2)((1/(t+1))−(1/(t+3)))dt =(1/2)∫_0 ^1 ((arctan(t+2))/(t+3))dt−(1/2)∫_0 ^1 ((arctan(t+2))/(t+1))dt =(1/2)H−(1/2)K H=∫_0 ^1 ((arctan(t+2))/(t+3))dt and K=∫_0 ^1 ((arctan(t+2))/(t+1))dt let f(x)=∫_0 ^1 ((arctan(t+x))/(t+3))dt with x≥0 ⇒ f^′ (x)=∫_0 ^1 (1/((t+3)(1+(t+x)^2 )))dt =∫_0 ^1 (dt/((t+3)(t^2 +2xt +x^2 +1))) let decompose F(t)=(1/((t+3)(t^2 +2xt +x^2 +1))) F(t)=(a/(t+3)) +((bt +c)/(t^2 +2xt +x^2 +1)) a =(1/(10−6x +x^2 )) lim_(t→+∞) tF(t)=0=a+b ⇒b=((−1)/(x^2 −6x +10)) F(0)=(1/(3(x^2 +1))) =(a/3) +(c/(x^2 +1)) ⇒(1/3) =((x^2 +1)/3)a +c ⇒ 1=(x^2 +1)a +3c ⇒3c=1−(x^2 +1)×(1/(x^2 −6x +10)) ⇒c =(1/3)(1−((x^2 +1)/(x^2 −6x +10))) ∫_0 ^1 F(t)dt =a ∫_0 ^1 (dt/(t+3)) +(b/2) ∫_0 ^1 ((2t +2x−2x )/(t^2 +2xt +x^2 +1))dt =a[ln(t+3)]_0 ^1 +(b/2)ln(t^2 +2xt +x^2 +1)]_0 ^x +c ∫_0 ^1 (dt/(t^2 +2xt +x^2 +1)) ∫_0 ^1 (dt/(t^2 +2xt +x^2 +1)) =∫_0 ^1 (dt/((t+x)^2 +1)) =_(t+x=u) ∫_x ^(x+1) (du/(1+u^2 )) =arctan(x+1)−arctanx ⇒ f^′ (x)=a{ln((4/3))}+(b/2)(ln(4x^2 +1)−ln(x^2 +1)+c(arctan(x+1)−arctanx) ⇒f(x)=aln((4/3))x +(b/2)∫ln(((4x^2 +1)/(x^2 +1)))dx +c∫ (arctan(x+1)−arctan(x))dx +c ....be continued....$
$l e t I = \int_{2}^{3} \frac{a r c t a n (x)}{1 - x^{2}} d x =_{x = 2 + t} \int_{0}^{1} \frac{a r c t a n (2 + t)}{1 - {(2 + t)}^{2}} d t$ $= - \int_{0}^{1} \frac{a r c t a n (2 + t)}{{(t + 2)}^{2} - 1} d t = - \int_{0}^{1} \frac{a r c t a n (t + 2)}{(t + 1) (t + 3)} d t$ $= - \frac{1}{2} \int_{0}^{1} a r c t a n (t + 2) (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$ $= \frac{1}{2} \int_{0}^{1} \frac{a r c t a n (t + 2)}{t + 3} d t - \frac{1}{2} \int_{0}^{1} \frac{a r c t a n (t + 2)}{t + 1} d t$ $= \frac{1}{2} H - \frac{1}{2} K$ $H = \int_{0}^{1} \frac{a r c t a n (t + 2)}{t + 3} d t a n d K = \int_{0}^{1} \frac{a r c t a n (t + 2)}{t + 1} d t$ $l e t f (x) = \int_{0}^{1} \frac{a r c t a n (t + x)}{t + 3} d t w i t h x ⩾ 0 \Rightarrow$ $f^{^{'}} (x) = \int_{0}^{1} \frac{1}{(t + 3) (1 + {(t + x)}^{2})} d t = \int_{0}^{1} \frac{d t}{(t + 3) (t^{2} + 2 x t + x^{2} + 1)}$ $l e t d e c o m p o s e F (t) = \frac{1}{(t + 3) (t^{2} + 2 x t + x^{2} + 1)}$ $F (t) = \frac{a}{t + 3} + \frac{b t + c}{t^{2} + 2 x t + x^{2} + 1}$ $a = \frac{1}{10 - 6 x + x^{2}}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = \frac{- 1}{x^{2} - 6 x + 10}$ $F (0) = \frac{1}{3 (x^{2} + 1)} = \frac{a}{3} + \frac{c}{x^{2} + 1} \Rightarrow \frac{1}{3} = \frac{x^{2} + 1}{3} a + c \Rightarrow$ $1 = (x^{2} + 1) a + 3 c \Rightarrow 3 c = 1 - (x^{2} + 1) \times \frac{1}{x^{2} - 6 x + 10}$ $\Rightarrow c = \frac{1}{3} (1 - \frac{x^{2} + 1}{x^{2} - 6 x + 10})$ $\int_{0}^{1} F (t) d t = a \int_{0}^{1} \frac{d t}{t + 3} + \frac{b}{2} \int_{0}^{1} \frac{2 t + 2 x - 2 x}{t^{2} + 2 x t + x^{2} + 1} d t$ ${= a {[l n (t + 3)]}_{0}^{1} + \frac{b}{2} l n (t^{2} + 2 x t + x^{2} + 1)]}_{0}^{x} + c \int_{0}^{1} \frac{d t}{t^{2} + 2 x t + x^{2} + 1}$ $\int_{0}^{1} \frac{d t}{t^{2} + 2 x t + x^{2} + 1} = \int_{0}^{1} \frac{d t}{{(t + x)}^{2} + 1} =_{t + x = u} \int_{x}^{x + 1} \frac{d u}{1 + u^{2}}$ $= a r c t a n (x + 1) - a r c t a n x \Rightarrow$ $f^{^{'}} (x) = a {l n (\frac{4}{3})} + \frac{b}{2} (l n (4 x^{2} + 1) - l n (x^{2} + 1) + c (a r c t a n (x + 1) - a r c t a n x)$ $\Rightarrow f (x) = a l n (\frac{4}{3}) x + \frac{b}{2} \int l n (\frac{4 x^{2} + 1}{x^{2} + 1}) d x + c \int (a r c t a n (x + 1) - a r c t a n (x)) d x + c$ $. . . . b e c o n t i n u e d . . . .$
Commented by TawaTawa last updated on 01/Nov/19

$God bless you sir . Waiting for the rest sir$ $Thanks for your time sir .$
Commented by mathmax by abdo last updated on 01/Nov/19

$y o u a r e w e l c o m e .$
	Answered by mind is power last updated on 31/Oct/19
	$let u=tan^(−1) (x) ⇒x=tg(u)⇒dx=(1+tg^2 (u) )du ∫_2 ^3 ((tan^(−1) (x))/(1−x^2 ))dx=∫_(tan^− (2)) ^(tsn^− (3)) ((u.(1+tg^2 (u))du)/(1−tg^2 (u))) =∫−udu+2∫((udu)/(1−tg^2 (u))) −ln(u)+2∫((udu)/(1−(((e^(iu) −e^(−iu) )/(i(e^(iu) +e^(−iu) ))))^2 ))2∫((udu.)/(((e^(iu) +e^(−iu) )^2 +(e^(iu) −e^(−iu) )^2 )/((e^(iu) +e^(−iu) )^2 ))) =2∫((u(e^(2iu) +e^(−2iu) +2))/(2e^(2iu) +2e^(−2iu) ))du=∫udu+2∫((udu)/(e^(2iu) +e^(−2iu) )) =2∫((ue^(2iu) du)/(1+e^(4iu) ))=2∫((ue^(2iu) du)/((e^(2iu) +i)(e^(2iu) −i)))=∫((udu)/(e^(2iu) +i))+∫((udu)/(e^(2iu) −i)) =∫((udu)/((e^(iu) +e^(−i(π/4)) )(e^(iu) −e^((−iπ)/4) )))+∫((udu)/((e^(iu) +e^(i(π/4)) )(e^(iu) −e^((iπ)/4) ))) =(1/(2e^((−iπ)/4) )){∫((udu)/(e^(iu) −e^((−iπ)/4) ))−∫((udu)/(e^(iu) +e^((−iπ)/4) ))}+(1/(2e^((iπ)/4) )){∫((udu)/(e^(iu) −e^((iπ)/4) ))−∫((udu)/(e^(iu) +e^((iπ)/4) ))} =(e^(i(π/2)) /2){∫(u/(e^(i(u+(π/4))) −1))du}−(e^(−i(π/2)) /2){∫((udu)/(e^(i(u−((3π)/4))) −1))}+(e^(−i(π/2)) /2){∫((udu)/(e^(i(u−(π/4))) −1))}−(e^(i(π/2)) /2)∫((udu)/(e^(i(u+((3π)/4))) −1)) ∫_1 ^(1−z) ((ln(t))/(1−t))=Li_2 (z) ∫_(tan^− (2)) ^(tan^− (3)) ((udu)/(e^(i(u+c)) −1))du w=e^(i(u+c)) ⇒u=−iln(w)−c⇒du=((−idw)/w) ⇒∫_e^(i(tan^− (2)+c)) ^e^(i(tan^− (3)+c)) ((−iln(w)−c)/(w−1)).−i(dw/w) =∫((−ln(w)+ic)/(w(w−1)))=∫((−ln(w)+ic)/(w−1))+∫((ln(w))/w)dw−∫((icdw)/w) =−Li_2 (1−e^(i(tan^− (3)+c)) )+li_2 (1−e^(i(tan^(−2) +c)) )+iclog((e^(i(tan^− (3)+c)) /e^(i(tan^− (2)+c)) ))+log(((itan^− (3)+ic)/(itan^− (2)+ic)))−ic(itan^− (3)−itan^− (2)) put this for c=i(π/4),((−3iπ)/4),((−iπ)/4),((3iπ)/4)$
	$let u = \tan^{- 1} (x)$ $\Rightarrow x = tg (u) \Rightarrow dx = (1 + {tg}^{2} (u)) du$ $\int_{2}^{3} \frac{\tan^{- 1} (x)}{1 - x^{2}} d x = \int_{\tan^{-} (2)}^{{tsn}^{-} (3)} \frac{u . (1 + {tg}^{2} (u)) du}{1 - {tg}^{2} (u)}$ $= \int - udu + 2 \int \frac{udu}{1 - {tg}^{2} (u)}$ $- \ln (u) + 2 \int \frac{udu}{1 - {(\frac{e^{iu} - e^{- iu}}{i (e^{iu} + e^{- iu})})}^{2}} 2 \int \frac{udu .}{\frac{{(e^{iu} + e^{- iu})}^{2} + {(e^{iu} - e^{- iu})}^{2}}{{(e^{iu} + e^{- iu})}^{2}}}$ $= 2 \int \frac{u (e^{2 iu} + e^{- 2 iu} + 2)}{{2 e}^{2 iu} + {2 e}^{- 2 iu}} du = \int udu + 2 \int \frac{udu}{e^{2 iu} + e^{- 2 iu}}$ $= 2 \int \frac{{ue}^{2 iu} du}{1 + e^{4 iu}} = 2 \int \frac{{ue}^{2 iu} du}{(e^{2 iu} + i) (e^{2 iu} - i)} = \int \frac{udu}{e^{2 iu} + i} + \int \frac{udu}{e^{2 iu} - i}$ $= \int \frac{udu}{(e^{iu} + e^{- i \frac{π}{4}}) (e^{iu} - e^{\frac{- i π}{4}})} + \int \frac{udu}{(e^{iu} + e^{i \frac{π}{4}}) (e^{iu} - e^{\frac{i π}{4}})}$ $= \frac{1}{{2 e}^{\frac{- i π}{4}}} {\int \frac{udu}{e^{iu} - e^{\frac{- i π}{4}}} - \int \frac{udu}{e^{iu} + e^{\frac{- i π}{4}}}} + \frac{1}{{2 e}^{\frac{i π}{4}}} {\int \frac{udu}{e^{iu} - e^{\frac{i π}{4}}} - \int \frac{udu}{e^{iu} + e^{\frac{i π}{4}}}}$ $= \frac{e^{i \frac{π}{2}}}{2} {\int \frac{u}{e^{i (u + \frac{π}{4})} - 1} du} - \frac{e^{- i \frac{π}{2}}}{2} {\int \frac{udu}{e^{i (u - \frac{3 π}{4})} - 1}} + \frac{e^{- i \frac{π}{2}}}{2} {\int \frac{udu}{e^{i (u - \frac{π}{4})} - 1}} - \frac{e^{i \frac{π}{2}}}{2} \int \frac{udu}{e^{i (u + \frac{3 π}{4})} - 1}$ $\int_{1}^{1 - z} \frac{\ln (t)}{1 - t} = {Li}_{2} (z)$ $\int_{\tan^{-} (2)}^{\tan^{-} (3)} \frac{udu}{e^{i (u + c)} - 1} du$ $w = e^{i (u + c)} \Rightarrow u = - iln (w) - c \Rightarrow du = \frac{- idw}{w}$ $\Rightarrow \int_{e^{i (\tan^{-} (2) + c)}}^{e^{i (\tan^{-} (3) + c)}} \frac{- iln (w) - c}{w - 1} . - i \frac{dw}{w}$ $= \int \frac{- \ln (w) + ic}{w (w - 1)} = \int \frac{- \ln (w) + ic}{w - 1} + \int \frac{\ln (w)}{w} dw - \int \frac{icdw}{w}$ $= - {Li}_{2} (1 - e^{i (\tan^{-} (3) + c)}) + {li}_{2} (1 - e^{i (\tan^{- 2} + c)}) + iclog (\frac{e^{i (\tan^{-} (3) + c)}}{e^{i (\tan^{-} (2) + c)}}) + \log (\frac{{itan}^{-} (3) + ic}{{itan}^{-} (2) + ic}) - ic ({itan}^{-} (3) - {itan}^{-} (2))$ $put this for c = i \frac{π}{4}, \frac{- 3 i π}{4}, \frac{- i π}{4}, \frac{3 i π}{4}$
	Commented by TawaTawa last updated on 31/Oct/19

	$Wow, God bless you sir . I appreciate your time sir$
	Commented by mind is power last updated on 31/Oct/19

	$most Welcom$
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