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Question Number 66906 by Kunal12588 last updated on 20/Aug/19

Intergrate I=∫ (t^2 /(1+t^4 )) dt

$${Intergrate}\:{I}=\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$

Commented by Prithwish sen last updated on 20/Aug/19

(1/(2i))∫(1/(1−it^2 )) − (1/(1+it^2 )) dt=(1/2)[−sin^(−1) it+ tan^(−1) it]+C please check

$$\frac{\mathrm{1}}{\mathrm{2i}}\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{it}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{it}^{\mathrm{2}} }\:\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left[−\mathrm{sin}^{−\mathrm{1}} \mathrm{it}+\:\mathrm{tan}^{−\mathrm{1}} \mathrm{it}\right]+\mathrm{C} \\ $$$$\mathrm{please}\:\mathrm{check} \\ $$

Commented by mathmax by abdo last updated on 21/Aug/19

I =∫ (t^2 /(1+t^4 ))dt =∫ (t^2 /((1−it^2 )(1+it^2 )))dt =(1/(2it^2 ))∫ t^2 ((1/(1−it^2 ))−(1/(1+it^2 )))dt =(1/(2i)) {∫ (dt/(1−it^2 ))−∫ (dt/(1+it^2 ))} but ∫ (dt/(1−it^2 )) =∫ (dt/((1−(√i)t)(1+(√i)t))) =(1/2)∫ {(1/(1−(√i)t)))+(1/(1+(√i)t))}dt =(1/(2(√i)))ln(((1+(√i)t)/(1−(√i)t))) +c_1 ∫ (dt/(1+it^2 )) =∫ (dt/(1+((√i)t)^2 )) =(1/(√i)) arctan((√i)t) +c_2 ⇒ I =(1/(4i(√i)))ln(((1+(√i)t)/(1−(√i)t)))−(1/(2i(√i))) arctan((√i)t)+C with (√i)=e^((iπ)/4)

$${I}\:=\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\int\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}−{it}^{\mathrm{2}} \right)\left(\mathrm{1}+{it}^{\mathrm{2}} \right)}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}{it}^{\mathrm{2}} }\int\:{t}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}−{it}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{it}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left\{\int\:\:\:\frac{{dt}}{\mathrm{1}−{it}^{\mathrm{2}} }−\int\:\:\frac{{dt}}{\mathrm{1}+{it}^{\mathrm{2}} }\right\}\:{but}\:\int\:\frac{{dt}}{\mathrm{1}−{it}^{\mathrm{2}} }\:=\int\:\:\frac{{dt}}{\left(\mathrm{1}−\sqrt{{i}}{t}\right)\left(\mathrm{1}+\sqrt{{i}}{t}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\left\{\frac{\mathrm{1}}{\left.\mathrm{1}−\sqrt{{i}}{t}\right)}+\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{i}}{t}}\right\}{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}{ln}\left(\frac{\mathrm{1}+\sqrt{{i}}{t}}{\mathrm{1}−\sqrt{{i}}{t}}\right)\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{dt}}{\mathrm{1}+{it}^{\mathrm{2}} }\:=\int\:\frac{{dt}}{\mathrm{1}+\left(\sqrt{{i}}{t}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\sqrt{{i}}}\:{arctan}\left(\sqrt{{i}}{t}\right)\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{4}{i}\sqrt{{i}}}{ln}\left(\frac{\mathrm{1}+\sqrt{{i}}{t}}{\mathrm{1}−\sqrt{{i}}{t}}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{i}}}\:{arctan}\left(\sqrt{{i}}{t}\right)+{C}\:\:\:\:{with}\:\sqrt{{i}}={e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$

Commented by Kunal12588 last updated on 21/Aug/19

thanks sir

$${thanks}\:{sir} \\ $$

Commented by turbo msup by abdo last updated on 21/Aug/19

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Commented by mathmax by abdo last updated on 26/Aug/19

classic method let decompose F(t) =(t^2 /(t^4 +1)) ⇒ F(t) =(t^2 /((t^2 +1)^2 −2t^2 )) =(t^2 /((t^2 +1−(√2)t)(t^2 +1+(√2)t))) =(t^2 /((t^2 +(√2)t +1)(t^2 −(√2)t +1))) =((at+b)/(t^2 +(√2)t +1)) +((ct+d)/(t^2 −(√2)t +1)) F(−t)=F(t)⇒((−at+b)/(t^2 −(√2)t+1))+((−ct +d)/(t^2 +(√2)t +1)) =F(t) ⇒c=−a and d=b ⇒F(t) =((at+b)/(t^2 +(√2)t +1)) +((−at+b)/(t^2 −(√2)t +1)) F(0) =0 =2b ⇒b=0 F(1) =(1/2) =(a/(2+(√2)))−(a/(2−(√2))) =(((2−(√2))a−(2+(√2))a)/2) ⇒ −2(√2)a =1 ⇒a =−(1/(2(√2))) ⇒ F(t) =−(1/(2(√2)))×(1/(t^2 +(√2)t +1)) +(1/(2(√2)))×(1/(t^2 −(√2)t +1)) ⇒ ∫ (t^2 /(1+t^4 ))dt =(1/(2(√2))) ∫ (dt/(t^2 −(√2)t +1))−(1/(2(√2)))∫ (dt/(t^2 +(√2)t +1)) but ∫ (dt/(t^2 +(√2)t +1)) =∫ (dt/(t^2 +2((√2)/2)t +(1/2)+(1/2))) =∫ (dt/((t+(1/(√2)))^2 +(1/2))) =_(t+(1/(√2))=(1/(√2))u) 2∫ (1/(1+u^2 )) (du/(√2)) =(1/(√2)) arctan(t(√2)+1) +c_1 ∫ (dt/(t^2 −(√2)t +1)) =_(t=−u) ∫ ((−du)/(u^2 +(√2)u +1)) =−(1/(√2))arctan(−t(√2) +1) +c_2 =(1/(√2))arctan(t(√2)−1)+c_2 ⇒ ∫ (t^2 /(1+t^4 ))dt =(1/(2(√2)))×(1/(√2)) arctan(t(√2)−1)−(1/(2(√2)))×(1/(√2)) arctan(t(√2) +1) +C =(1/4) arctan(t(√2)−1)−(1/4) arctan(t(√2) +1) +C

$${classic}\:{method}\:\:\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} }\:=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{2}}{t}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{2}}{t}\right)} \\ $$$$=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)}\:=\frac{{at}+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\Rightarrow\frac{−{at}+{b}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\frac{−{ct}\:+{d}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:={F}\left({t}\right)\:\Rightarrow{c}=−{a}\:{and}\:{d}={b} \\ $$$$\Rightarrow{F}\left({t}\right)\:=\frac{{at}+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{at}+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{0} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{a}}{\mathrm{2}+\sqrt{\mathrm{2}}}−\frac{{a}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}}{\mathrm{2}}\:\Rightarrow \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}{a}\:=\mathrm{1}\:\Rightarrow{a}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:\:{but} \\ $$$$\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{t}\:\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\int\:\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{t}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{u}} \:\:\:\:\mathrm{2}\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\sqrt{\mathrm{2}}}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left({t}\sqrt{\mathrm{2}}+\mathrm{1}\right)\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=_{{t}=−{u}} \:\:\:\:\:\:\int\:\:\frac{−{du}}{{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{arctan}\left(−{t}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{arctan}\left({t}\sqrt{\mathrm{2}}−\mathrm{1}\right)+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left({t}\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left({t}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}\left({t}\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}\left({t}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\:+{C} \\ $$$$ \\ $$

Answered by Tanmay chaudhury last updated on 20/Aug/19

∫(dt/(t^2 +(1/t^2 ))) (1/2)∫((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt (1/2)∫((d(t+(1/t)))/((t+(1/t))^2 −2))+(1/2)∫((d(t−(1/t)))/((t−(1/t))^2 +2)) formula ∫(dx/(x^2 −a^2 ))=(1/(2a))ln(((x−a)/(x+a)))+c ∫(dx/(x^2 +a^2 ))=(1/a)tan^(−1) ((x/a))+c (1/2)×(1/(2(√2)))ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))+(1/2)×(1/(√2))tan^(−1) (((t−(1/t))/(√2)))+c

$$\int\frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$${formula}\:\int\frac{{dx}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{a}}{ln}\left(\frac{{x}−{a}}{{x}+{a}}\right)+{c} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$ \\ $$

Commented by Kunal12588 last updated on 21/Aug/19

thanks sir

$${thanks}\:{sir} \\ $$

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