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Question Number 34227 by abdo imad last updated on 03/May/18
calculate ∫_0 ^1 arctan(x^2 )dx
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{2}} \right){dx}\: \\ $$
Commented by math khazana by abdo last updated on 04/May/18
let put I = ∫_0 ^1 arctan(x^2 )dx by parts u^′ =1 and v=arctan(x^2 ) ⇒ I = [x arctan(x^2 )]_0 ^1 −∫_0 ^1 x((2x)/(1+x^4 ))dx =(π/4) −2 ∫_0 ^1 (x^2 /(1+x^4 ))dx let decompose F(x)= (x^2 /(1+x^4 )) = (x^2 /((x^2 +1)^2 −2x^2 )) = (x^2 /((x^2 −(√2)x+1)(x^2 +(√2)x +1))) F(x)= ((ax+b)/(x^2 −(√2)x+1 )) +((cx+d)/(x^2 +(√2)x +1)) F(−x) =F(x) ⇒ c=−a and d=b ⇒ F(x) = ((ax +b)/(x^2 −(√2)x +1)) +((−ax+b)/(x^2 +(√2)x +1)) F(0)=0 =2b ⇒b=0 ⇒ F(x)= ((ax)/(x^2 −(√2)x +1)) −((ax)/(x^2 +(√2)x+1)) F(1) =(1/2) = (a/(2−(√2))) −(a/(2+(√2))) =(((2+(√2)−2+(√2))/2))a =(√2) a ⇒a= (1/(2(√2))) ⇒ F(x)= (1/(2(√2)))( (x/(x^2 −(√2)x+1)) −(x/(x^2 +(√2)x+1))) ∫_0 ^1 (x^2 /(1+x^4 ))dx = (1/(2(√2))) ∫_0 ^1 (x/(x^2 −(√2)x+1))dx −(1/(2(√2)))∫_0 ^1 (x/(x^2 +(√2)x+1))dx
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{2}} \right){dx}\:{by}\:{parts}\:{u}^{'} =\mathrm{1}\:{and}\: \\ $$$${v}={arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow\:{I}\:=\:\left[{x}\:{arctan}\left({x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} }\:=\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}+\mathrm{1}\:}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(−{x}\right)\:={F}\left({x}\right)\:\Rightarrow\:{c}=−{a}\:{and}\:{d}={b}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{{ax}\:+{b}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:+\frac{−{ax}+{b}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}\:=\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\:\frac{{ax}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:−\frac{{ax}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{{a}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:−\frac{{a}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:=\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}−\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right){a} \\ $$$$=\sqrt{\mathrm{2}}\:{a}\:\Rightarrow{a}=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\:\:\frac{{x}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:−\frac{{x}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx} \\ $$
Commented by math khazana by abdo last updated on 04/May/18
∫_0 ^1 (x/(x^2 −(√2)x +1))dx = (1/2)∫_0 ^1 ((2x−(√2) +(√2))/(x^2 −(√2)x +1))dx =(1/2)[ln(x^2 −(√2)x+1)]_0 ^1 +((√2)/2) ∫_0 ^1 (dx/(x^2 −(√2)x +1)) =(1/2)ln(2−(√2)) +((√2)/2) ∫_0 ^1 (dx/(x^2 −(√2) x+1)) but ∫_0 ^1 (dx/(x^2 −(√2) x+1)) = ∫_0 ^1 (dx/(x^2 −2((√2)/2)x +(1/4) +(3/4))) = ∫_0 ^1 (dx/((x−((√2)/2))^2 +(3/4))) =_(x−((√2)/2)=((√3)/2)t) ∫_(−((√2)/( (√3)))) ^((2/( (√3)))(1−((√2)/2))) ((((√3)/2)dt)/((3/4)(1+t^2 ))) =(4/3) ((√3)/2) ∫_(−((√2)/( (√3)))) ^((2/( (√3))) −((√2)/( (√3)))) (dt/(1+t^2 )) = ((2(√3))/3) (arctan((2/( (√3)))−((√2)/( (√3)))) +arctan(((√2)/( (√3)))))⇒ ∫_0 ^1 (x/(x^2 −(√2)x +1))dx =(1/2)ln(2−(√2)) +((2(√3))/3) ( arctan((2/( (√3))) −((√2)/( (√3)))) +arctan(((√2)/( (√3))))) ∫_0 ^1 (x/(x^2 +(√2)x +1))dx =_(x=−t) ∫_0 ^(−1) ((tdt)/(t^2 −(√2) t +1)) and this caculated above....
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{x}+\mathrm{1}}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{x}+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\:\int_{−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \:\:\:\:\:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}}{\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left({arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\:+{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right) \\ $$$$+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left(\:{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\:+{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right)\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx}\:=_{{x}=−{t}} \:\:\int_{\mathrm{0}} ^{−\mathrm{1}} \:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}} \\ $$$${and}\:{this}\:{caculated}\:{above}…. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18
let I=∫tan^(−1) (x^2 )dx =tan^(−1) (x^2 )∫dx−∫[(d/dx)tan^(−1) x^(2.) .∫dx]dx =xtan^(−1 ) (x^2 )−∫((2x)/(1+x^4 )).xdx =xtan^(−1) x−2∫(x^2 /(1+x^4 ))dx i have alredy solved ∫(x^2 /(1+x^4 ))dx pls co−relate
$${let}\:{I}=\int{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right){dx} \\ $$$$={tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)\int{dx}−\int\left[\frac{{d}}{{dx}}{tan}^{−\mathrm{1}} {x}^{\mathrm{2}.} .\int{dx}\right]{dx} \\ $$$$={xtan}^{−\mathrm{1}\:} \left({x}^{\mathrm{2}} \right)−\int\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} \:}.{xdx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\mathrm{2}\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${i}\:{have}\:{alredy}\:{solved}\:\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${pls}\:{co}−{relate} \\ $$

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