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Question Number 104302 by bobhans last updated on 20/Jul/20

Commented by bobhans last updated on 21/Jul/20
$by parts { ((u=arc tan ((1/(2x^2 ))) ⇒du =((−(1/x^3 ))/(1+(1/(4x^4 )))) dx)),((dv = dx ⇒v = x)) :} I= x arc tan ((1/(2x^2 )))+∫ ((4x^2 )/(4x^4 +1)) dx I= x arctan ((1/(2x^2 ))) +∫ (x/(2x^2 −2x+1))−(x/(2x^2 +2x+1)) dx$
$b y p a r t s$ ${\begin{cases} u = arc \tan (\frac{1}{2 x^{2}}) \Rightarrow d u = \frac{- \frac{1}{x^{3}}}{1 + \frac{1}{4 x^{4}}} d x \\ d v = d x \Rightarrow v = x \end{cases}$ $I = x arc \tan (\frac{1}{2 x^{2}}) + \int \frac{4 x^{2}}{4 x^{4} + 1} d x$ $I = x \arctan (\frac{1}{2 x^{2}}) + \int \frac{x}{2 x^{2} - 2 x + 1} - \frac{x}{2 x^{2} + 2 x + 1} d x$
Commented by john santu last updated on 20/Jul/20
$I= x tan^(−1) ((1/(2x^2 )))+∫(x/(2x^2 −2x+1))−(x/(2x^2 +2x+1))dx let J= ∫((2x)/(4x^2 −4x+2))−((2x)/(4x^2 +4x+2))dx J= ∫ ((2x)/((2x−1)^2 +1))−((2x)/((2x+1)^2 +1))dx J=∫((2x−1+1)/((2x−1)^2 +1))−((2x+1−1)/((2x+1)^2 +1))dx J=∫ [((2x−1)/((2x−1)^2 +1))+(1/((2x−1)^2 +1))−((2x+1)/((2x+1)^2 +1))+(1/((2x+1)^2 +1))]dx J= (1/2)ln ∣((2x^2 −2x+1)/(2x^2 +2x+1))∣ +(1/2){tan^(−1) (2x−1)+tan^(−1) (2x−1)} we conclude I= x tan^(−1) ((1/(2x^2 )))+(1/2)ln ∣((2x^2 −2x+1)/(2x^2 +2x+1))∣+ (1/2){tan^(−1) (2x−1)+tan^(−1) (2x+1)} + C (JS ⊛)$
$I = x \tan^{- 1} (\frac{1}{2 x^{2}}) + \int \frac{x}{2 x^{2} - 2 x + 1} - \frac{x}{2 x^{2} + 2 x + 1} d x$ $l e t J = \int \frac{2 x}{4 x^{2} - 4 x + 2} - \frac{2 x}{4 x^{2} + 4 x + 2} d x$ $J = \int \frac{2 x}{{(2 x - 1)}^{2} + 1} - \frac{2 x}{{(2 x + 1)}^{2} + 1} d x$ $J = \int \frac{2 x - 1 + 1}{{(2 x - 1)}^{2} + 1} - \frac{2 x + 1 - 1}{{(2 x + 1)}^{2} + 1} d x$ $J = \int [\frac{2 x - 1}{{(2 x - 1)}^{2} + 1} + \frac{1}{{(2 x - 1)}^{2} + 1} - \frac{2 x + 1}{{(2 x + 1)}^{2} + 1} + \frac{1}{{(2 x + 1)}^{2} + 1}] d x$ $J = \frac{1}{2} \ln ∣ \frac{2 x^{2} - 2 x + 1}{2 x^{2} + 2 x + 1} ∣ + \frac{1}{2} {\tan^{- 1} (2 x - 1) + \tan^{- 1} (2 x - 1)}$ $w e c o n c l u d e$ $I = x \tan^{- 1} (\frac{1}{2 x^{2}}) + \frac{1}{2} \ln ∣ \frac{2 x^{2} - 2 x + 1}{2 x^{2} + 2 x + 1} ∣ +$ $\frac{1}{2} {\tan^{- 1} (2 x - 1) + \tan^{- 1} (2 x + 1)} + C$ $(J S ⊛)$
	Answered by OlafThorendsen last updated on 20/Jul/20

	$\arctan u + \arctan \frac{1}{u} = \frac{π}{2}$ $I = \int \arctan (\frac{1}{2 x^{2}}) d x$ $I = \frac{π}{2} x - \int \arctan (2 x^{2}) d x$ $I = \frac{π}{2} x - x \arctan (2 x^{2}) + \int \frac{4 x^{2}}{1 + 4 x^{4}} d x$ $. . .$
	Commented by Ar Brandon last updated on 20/Jul/20
	$Cool !!! Mind if I continue... J=∫((4x^2 )/(1+4x^4 ))dx=∫{((2x^2 +1)/(4x^4 +1))+((2x^2 −1)/(4x^4 +1))}dx =∫{((2+(1/x^2 ))/(4x^2 +(1/x^2 )))+((2−(1/x^2 ))/(4x^2 +(1/x^2 )))}dx=∫{((2+(1/x^2 ))/((2x−(1/x))^2 +4))+((2−(1/x^2 ))/((2x+(1/x))^2 −4))}dx =(1/2)Arctan[((2x^2 −1)/(2x))]−(1/2)Arctanh[((2x^2 +1)/(2x))]+C$
	$Cool!!! Mind if I continue . . .$ $J = \int \frac{4 x^{2}}{1 + 4 x^{4}} d x = \int {\frac{2 x^{2} + 1}{4 x^{4} + 1} + \frac{2 x^{2} - 1}{4 x^{4} + 1}} d x$ $= \int {\frac{2 + \frac{1}{x^{2}}}{4 x^{2} + \frac{1}{x^{2}}} + \frac{2 - \frac{1}{x^{2}}}{4 x^{2} + \frac{1}{x^{2}}}} d x = \int {\frac{2 + \frac{1}{x^{2}}}{{(2 x - \frac{1}{x})}^{2} + 4} + \frac{2 - \frac{1}{x^{2}}}{{(2 x + \frac{1}{x})}^{2} - 4}} d x$ $= \frac{1}{2} Arctan [\frac{2 x^{2} - 1}{2 x}] - \frac{1}{2} Arctanh [\frac{2 x^{2} + 1}{2 x}] + C$
	Answered by Dwaipayan Shikari last updated on 20/Jul/20
	$∫tan^(−1) ((1/(2x^2 )))dx=∫tan^(−1) (2x+1)−tan^(−1) (2x−1)dx (1/2)∫tan^(−1) u−(1/2)∫tan^(−1) p {take 2x+1=t 2x−1=p (1/2)u tan^(−1) u−(1/2)∫(u/(u^2 +1))−(1/2)p tan^(−1) p−p+(1/2)∫(p/(p^2 +1)) (1/2)(2x+1)tan^(−1) (2x+1)−(1/4)log((2x+1)^2 +1)−(1/2)(2x−1)tan^(−1) (2x−1) +(1/4)log((2x−1)^2 +1)+C$
	$\int \tan^{- 1} (\frac{1}{{2 x}^{2}}) dx = \int \tan^{- 1} (2 x + 1) - \tan^{- 1} (2 x - 1) dx$ $\frac{1}{2} \int \tan^{- 1} u - \frac{1}{2} \int \tan^{- 1} p {take 2 x + 1 = t 2 x - 1 = p$ $\frac{1}{2} u \tan^{- 1} u - \frac{1}{2} \int \frac{u}{u^{2} + 1} - \frac{1}{2} p \tan^{- 1} p - p + \frac{1}{2} \int \frac{p}{p^{2} + 1}$ $\frac{1}{2} (2 x + 1) \tan^{- 1} (2 x + 1) - \frac{1}{4} \log ({(2 x + 1)}^{2} + 1) - \frac{1}{2} (2 x - 1) \tan^{- 1} (2 x - 1)$ $+ \frac{1}{4} \log ({(2 x - 1)}^{2} + 1) + C$
	Commented by Ar Brandon last updated on 20/Jul/20
	wow ! that was creative ��
	Commented by Dwaipayan Shikari last updated on 20/Jul/20
	Sir olaftheorendsen , gives me this idea. ��
	Answered by mathmax by abdo last updated on 21/Jul/20
	$A =∫ arctan((1/(2x^2 )))dx ⇒A= ∫((π/2)−arctan(2x^2 ))dx A =((πx)/2) −∫ arctan(2x^2 )dx by parts u^′ =1 and v =arctan(2x^2 ) ⇒ ∫ arctan(2x^2 )dx =x arctan(2x^2 )−∫ x.((4x)/(1+4x^4 ))dx =xarctan(2x^2 )−4 ∫ (x^2 /(1+4x^4 )) dx we have ∫ (x^2 /(4x^4 +1))dx =∫ (x^2 /(((√2)x)^4 +1)) =_((√2)x =t) ∫ (t^2 /(2(t^4 +1))) (dt/(√2)) =(1/(2(√2)))∫ (t^2 /(t^4 +1))dt =(1/(2(√2)))∫ (1/(t^2 +(1/t^2 )))dt =(1/(4(√2))) ∫ ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 ))) dt =(1/(4(√2))) ∫ ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt(→u =t+(1/t))+(1/(4(√2))) ∫ ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt(→v =t−(1/t)) =(1/(4(√2))) ∫ (du/(u^2 −2)) +(1/(4(√2))) ∫ (dv/(v^2 +2)) we have ∫ (du/(u^2 −2)) =(1/(2(√2)))∫((1/(u−(√2)))−(1/(u+(√2))))du =(1/(2(√2)))ln∣((u−(√2))/(u+(√2)))∣+c_1 =(1/(2(√2)))ln∣((t+(1/t)−(√2))/(t+(1/t)+(√2)))∣+c_1 =(1/(2(√2)))ln∣(((√2)x+(1/((√2)x))−(√2))/((√2)x+(1/((√2)x))+(√2)))∣ +c_1 ∫ (dv/(v^2 +2)) =_(v=(√2)z) ∫ (((√2)dz)/(2(1+z^2 ))) =((√2)/2) arctanz +c_2 =((√2)/2) arctan((v/(√2))) +c_2 =((√2)/2)arctan((1/(√2)){(√2)x−(1/((√2)x))}) +c_2 ⇒ A =x arctan(2x^2 )−(1/(√2)){(1/(2(√2)))ln∣((2x^2 +1−2x)/(2x^2 +1+2x))∣ +((√2)/2) arctan((1/(√2))((√2)x−(1/((√2)x)))) +C A =xarctan(2x^2 )−(1/4)ln∣((2x^2 −2x+1)/(2x^2 +2x+1))∣−(1/2) arctan{x−(1/(2x))} +C$
	$A = \int \arctan (\frac{1}{{2 x}^{2}}) dx \Rightarrow A = \int (\frac{π}{2} - \arctan ({2 x}^{2})) dx$ $A = \frac{π x}{2} - \int \arctan ({2 x}^{2}) dx by parts u^{^{'}} = 1 and v = \arctan ({2 x}^{2}) \Rightarrow$ $\int \arctan ({2 x}^{2}) dx = x \arctan ({2 x}^{2}) - \int x . \frac{4 x}{1 + {4 x}^{4}} dx$ $= xarctan ({2 x}^{2}) - 4 \int \frac{x^{2}}{1 + {4 x}^{4}} dx we have$ $\int \frac{x^{2}}{{4 x}^{4} + 1} dx = \int \frac{x^{2}}{{(\sqrt{2} x)}^{4} + 1} =_{\sqrt{2} x = t} \int \frac{t^{2}}{2 (t^{4} + 1)} \frac{dt}{\sqrt{2}}$ $= \frac{1}{2 \sqrt{2}} \int \frac{t^{2}}{t^{4} + 1} dt = \frac{1}{2 \sqrt{2}} \int \frac{1}{t^{2} + \frac{1}{t^{2}}} dt = \frac{1}{4 \sqrt{2}} \int \frac{1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt$ $= \frac{1}{4 \sqrt{2}} \int \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt (\to u = t + \frac{1}{t}) + \frac{1}{4 \sqrt{2}} \int \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} dt (\to v = t - \frac{1}{t})$ $= \frac{1}{4 \sqrt{2}} \int \frac{du}{u^{2} - 2} + \frac{1}{4 \sqrt{2}} \int \frac{dv}{v^{2} + 2} we have$ $\int \frac{du}{u^{2} - 2} = \frac{1}{2 \sqrt{2}} \int (\frac{1}{u - \sqrt{2}} - \frac{1}{u + \sqrt{2}}) du = \frac{1}{2 \sqrt{2}} \ln ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣ + c_{1} = \frac{1}{2 \sqrt{2}} \ln ∣ \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} ∣ + c_{1}$ $= \frac{1}{2 \sqrt{2}} \ln ∣ \frac{\sqrt{2} x + \frac{1}{\sqrt{2} x} - \sqrt{2}}{\sqrt{2} x + \frac{1}{\sqrt{2} x} + \sqrt{2}} ∣ + c_{1}$ $\int \frac{dv}{v^{2} + 2} =_{v = \sqrt{2} z} \int \frac{\sqrt{2} dz}{2 (1 + z^{2})} = \frac{\sqrt{2}}{2} arctanz + c_{2} = \frac{\sqrt{2}}{2} \arctan (\frac{v}{\sqrt{2}}) + c_{2}$ $= \frac{\sqrt{2}}{2} \arctan (\frac{1}{\sqrt{2}} {\sqrt{2} x - \frac{1}{\sqrt{2} x}}) + c_{2} \Rightarrow$ $A = x \arctan ({2 x}^{2}) - \frac{1}{\sqrt{2}} {\frac{1}{2 \sqrt{2}} \ln ∣ \frac{{2 x}^{2} + 1 - 2 x}{{2 x}^{2} + 1 + 2 x} ∣ + \frac{\sqrt{2}}{2} \arctan (\frac{1}{\sqrt{2}} (\sqrt{2} x - \frac{1}{\sqrt{2} x})) + C$ $A = xarctan ({2 x}^{2}) - \frac{1}{4} \ln ∣ \frac{{2 x}^{2} - 2 x + 1}{{2 x}^{2} + 2 x + 1} ∣ - \frac{1}{2} \arctan {x - \frac{1}{2 x}} + C$
	Commented by mathmax by abdo last updated on 21/Jul/20

	$sorry A = \frac{π x}{2} - \int \arctan ({2 x}^{2}) dx \Rightarrow$ $A = \frac{π x}{2} - x \arctan ({2 x}^{2}) + \frac{1}{4} \ln ∣ \frac{{2 x}^{2} - 2 x + 1}{{2 x}^{2} + 2 x + 1} ∣ + \frac{1}{2} \arctan (x - \frac{1}{2 x}) + C$
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