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Question Number 35061 by math khazana by abdo last updated on 14/May/18

find ∫_0 ^∞ ((x^2 +3)/((x^4 +1)^2 ))dx

$${find}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Commented by math khazana by abdo last updated on 15/May/18

let put I = ∫_0 ^∞ ((x^2 +3)/((x^4 +1)^2 ))dx 2I =∫_(−∞) ^(+∞) ((x^2 +3)/((x^4 +1)^2 ))dx let consider the complex function f(z) = ((z^2 +3)/((z^4 +1)^2 )) f(z) = ((z^2 +3)/((z^2 −i)^2 (z^2 +i)^2 )) = ((z^2 +3)/((z−(√i))^2 (z+(√i))^2 (z−(√(−i)))^2 (z +(√(−i)))^2 )) = ((z^2 +3)/((z−e^(i(π/4)) )^2 (z +e^(i(π/4)) )^2 (z −e^(−i(π/4)) )^2 (z +e^(−i(π/4)) )^2 )) all poles of f are doubles ∫_(−∞) ^(+∞) f(z)dz =2iπ{ Res(f ,e^(i(π/4)) ) +Res(f, −e^(−i(π/4)) )} Res(f ,e^(i(π/4)) ) =lim_(z→e^(i(π/4)) ) (1/((2−1)!)) { (z−e^(i(π/4)) )^2 f(z)}^′ =lim_(z→e^(i(π/4)) ) ( ((z^2 +3)/((z +(√i))^2 (z^2 +i)^2 )))^′ = lim_(z→e^(i(π/4)) ) ((2z (z +(√i))^2 (z^2 +i)^2 − (z^2 +3)( 2(z+(√i))(z^2 +i)^2 +4z(z^2 +i)(z+(√i))^2 ))/((z+(√i))^4 (z^2 +i)^4 )) =lim_(z→e^(i(π/4)) ) ((2z(z+(√i))(z^2 +i) −2(z^2 +3)(z^2 +i +4z(z+(√i))))/((z +(√i))^3 (z^2 +i)^3 )) = ((2e^(i(π/4)) 2e^(i(π/4)) (2i) −2(i+3)(2i +4e^(i(π/4)) (2e^(i(π/4)) ))/((2e^(i(π/4)) )^2 (2i)^3 )) =((−8 −2(i+3)(10i))/(−8i .4i)) = ((8 +20i(i+3))/(−32)) = −((8 −20 +60i)/(32)) =−((−12 +60i)/(32)) =((−3 +15i)/8) for Res(f ,−e^(−i(π/4)) ) wefolow the same method... be continued...

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{2}{I}\:\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:{f}\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{3}}{\left({z}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${f}\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} +\mathrm{3}}{\left({z}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{3}}{\left({z}−\sqrt{{i}}\right)^{\mathrm{2}} \left({z}+\sqrt{{i}}\right)^{\mathrm{2}} \left({z}−\sqrt{−{i}}\right)^{\mathrm{2}} \left({z}\:+\sqrt{−{i}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{3}}{\left({z}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} } \\ $$$${all}\:{poles}\:{of}\:{f}\:{are}\:{doubles} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({f}\:,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:+{Res}\left({f},\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({f}\:,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\:\left({z}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} {f}\left({z}\right)\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\:\:\:\left(\:\frac{{z}^{\mathrm{2}} \:+\mathrm{3}}{\left({z}\:+\sqrt{{i}}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} }\right)^{'} \:\:\: \\ $$$$=\:{lim}_{{z}\rightarrow{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{2}{z}\:\left({z}\:+\sqrt{{i}}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \:−\:\left({z}^{\mathrm{2}} +\mathrm{3}\right)\left(\:\mathrm{2}\left({z}+\sqrt{{i}}\right)\left({z}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} \:+\mathrm{4}{z}\left({z}^{\mathrm{2}} +{i}\right)\left({z}+\sqrt{{i}}\right)^{\mathrm{2}} \right)}{\left({z}+\sqrt{{i}}\right)^{\mathrm{4}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{2}{z}\left({z}+\sqrt{{i}}\right)\left({z}^{\mathrm{2}} \:+{i}\right)\:−\mathrm{2}\left({z}^{\mathrm{2}} +\mathrm{3}\right)\left({z}^{\mathrm{2}} \:+{i}\:+\mathrm{4}{z}\left({z}+\sqrt{{i}}\right)\right)}{\left({z}\:+\sqrt{{i}}\right)^{\mathrm{3}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)\:−\mathrm{2}\left({i}+\mathrm{3}\right)\left(\mathrm{2}{i}\:+\mathrm{4}{e}^{{i}\frac{\pi}{\mathrm{4}}} \left(\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right.}{\left(\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{8}\:−\mathrm{2}\left({i}+\mathrm{3}\right)\left(\mathrm{10}{i}\right)}{−\mathrm{8}{i}\:.\mathrm{4}{i}}\:=\:\frac{\mathrm{8}\:\:+\mathrm{20}{i}\left({i}+\mathrm{3}\right)}{−\mathrm{32}} \\ $$$$=\:−\frac{\mathrm{8}\:−\mathrm{20}\:+\mathrm{60}{i}}{\mathrm{32}}\:=−\frac{−\mathrm{12}\:+\mathrm{60}{i}}{\mathrm{32}}\:=\frac{−\mathrm{3}\:+\mathrm{15}{i}}{\mathrm{8}} \\ $$$${for}\:{Res}\left({f}\:,−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:\:{wefolow}\:\:{the}\:{same}\:{method}... \\ $$$${be}\:{continued}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/May/18

=∫_0 ^∞ ((x^2 +3)/(x^4 (x^2 +(1/(x^2 )))^2 )) =∫_0 ^∞ ((1/x^(2 ) )/((x^2 +(1/(x^(2 ) )))^(2 ) ))+∫_0 ^∞ ((3/x^4 )/((x^2 +(1/(x^2 )))^2 )) I_1 =(1/2)∫_0 ^∞ (((1+(1/(x^2 )))−(1−(1/(x^2 ))))/((x^2 +(1/x^2 ) )^(2 ) ))dx =(1/2)∫_0 ^∞ (((1+(1/(x^2 ))) dx)/({(x−(1/x))^2 +2}^2 )) −(1/2)∫_0 ^∞ (((1−(1/(x^2 ))) dx)/({(x+(1/x))^2 −2}^2 )) contd z_1 =x−(1/x) z_(2 ) =x+(1/x) (1/2)∫_(−∞) ^∞ (dz_1 /((z_1 ^2 +2)^(2 ) )) −(1/2)∫_∞ ^∞ (dz_2 /((z_2 ^2 −2)^2 ))→its value 0 z_1 =(√2) tanθ (1/2)∫_(−Π/2) ^(Π/2) (((√2) sec^2 θ)/(2^2 sec^4 θ)) dθ ((√2)/8)∫_(−Π/2) ^(Π/2) ((1+cos2θ)/2)dθ =((√2)/(16))(Π)+(((√2) )/(16))×∣((sin2θ)/2)∣_(−Π/2) ^(Π/2) =(((√2)(Π))/(16))+(((√(2 )) )/(32)){sinΠ−sin(−Π)} =(√2) (Π/(16)) pls check

$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{4}} \left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} \:\:}\right)^{\mathrm{2}\:} }+\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{3}}{{x}^{\mathrm{4}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:\:}\right)^{\mathrm{2}} } \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:\:}\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\right)^{\mathrm{2}\:} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)\:{dx}}{\left\{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\right\}^{\mathrm{2}} \:}\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)\:{dx}}{\left\{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right\}^{\mathrm{2}} } \\ $$$${contd} \\ $$$${z}_{\mathrm{1}} ={x}−\frac{\mathrm{1}}{{x}}\:\:\:\:\:\:\:{z}_{\mathrm{2}\:} ={x}+\frac{\mathrm{1}}{{x}}\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{dz}_{\mathrm{1}} }{\left({z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}\:} }\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\infty} ^{\infty} \frac{{dz}_{\mathrm{2}} }{\left({z}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }\rightarrow{its}\:{value}\:\mathrm{0} \\ $$$${z}_{\mathrm{1}} =\sqrt{\mathrm{2}}\:{tan}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{−\Pi/\mathrm{2}} ^{\Pi/\mathrm{2}} \frac{\sqrt{\mathrm{2}}\:{sec}^{\mathrm{2}} \theta}{\mathrm{2}^{\mathrm{2}} {sec}^{\mathrm{4}} \theta}\:{d}\theta \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int_{−\Pi/\mathrm{2}} ^{\Pi/\mathrm{2}} \frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\left(\Pi\right)+\frac{\sqrt{\mathrm{2}}\:}{\mathrm{16}}×\mid\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\mid_{−\Pi/\mathrm{2}} ^{\Pi/\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{2}}\left(\Pi\right)}{\mathrm{16}}+\frac{\sqrt{\mathrm{2}\:}\:}{\mathrm{32}}\left\{{sin}\Pi−{sin}\left(−\Pi\right)\right\} \\ $$$$=\sqrt{\mathrm{2}}\:\frac{\Pi}{\mathrm{16}} \\ $$$${pls}\:{check} \\ $$$$ \\ $$

Answered by MJS last updated on 15/May/18

∫((x^2 +3)/((x^4 +1)^2 ))dx=i [((Ostrogradski′s Method)),((∫((p(x))/(q(x)))dx=((p_1 (x))/(q_1 (x)))+∫((p_2 (x))/(q_2 (x)))dx)),((q_1 (x)=gcd(q(x),q′(x)))),((q_2 (x)=((q(x))/(q_1 (x))))),((p(x)=x^2 +3; q(x)=(x^4 +1)^2 )),((q′(x)=8x^3 (x^4 +1); q_1 (x)=q_2 (x)=x^4 +1)),((now we calculate the factors of)),((p_1 (x) and p_2 (x) with degree(p_n )<degree(q_n ))),((using this equation (comparing factors))),((((p(x))/(q(x)))=(d/dx)[((p_1 (x))/(q_1 (x)))]+((p_2 (x))/(q_2 (x))))),((p_1 (x)=a_1 x^3 +b_1 x^2 +c_1 x+d_1 )),((p_2 (x)=a_2 x^3 +b_2 x^2 +c_2 x+d_2 )),((⇒ p_1 (x)=(1/4)(x^3 +3x); p_2 (x)=(1/4)(x^2 +9))) ] =((x(x^2 +3))/(4(x^4 +1)))+(1/4)∫((x^2 +9)/(x^4 +1))dx= ∫((x^2 +9)/(x^4 +1))dx=∫((x^2 +9)/((x^2 +(√2)x+1)(x^2 −(√2)x+1)))dx= =((√2)/4)∫(((8x+9(√2))/(x^2 +(√2)x+1))−((8x−9(√2))/(x^2 −(√2)x+1)))dx= =((√2)/4)∫(4((2x+(√2))/(x^2 +(√2)x+1))+5(√2)(1/(x^2 +(√2)x+1))−4((2x−(√2))/(x^2 −(√2)x+1))+5(√2)(1/(x^2 −(√2)x+1)))dx= [((the 1^(st) and 3^(rd) terms are of the form)),((F(x)=∫a((f′(x))/(f(x)))dx ⇒ F(x)=aln(f(x)))),((the 2^(nd) and 4^(th) terms can be solved using)),((u=(√2)x±1 → dx=(du/(√2)) and lead to)),((F(x)=(√2)∫(du/(u^2 +2)) ⇒ F(x)=(√2)arctan(u))) ] =(√2)(ln(x^2 +(√2)x+1)−ln(x^2 −(√2)x+1))+((5(√2))/2)(arctan((√2)x+1)+arctan((√2)x−1)) =((x(x^2 +3))/(4(x^4 +1)))+((√2)/4)ln(((∣x^2 +(√2)x+1∣)/(∣x^2 −(√2)x+1∣)))+((5(√2))/8)(arctan((√2)x+1)+arctan((√2)x−1))+C ∫_0 ^∞ ((x^2 +3)/((x^4 +1)^2 ))dx=((5(√2))/8)π

$$\int\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}={i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}}\\{\int\frac{{p}\left({x}\right)}{{q}\left({x}\right)}{dx}=\frac{{p}_{\mathrm{1}} \left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{p}_{\mathrm{2}} \left({x}\right)}{{q}_{\mathrm{2}} \left({x}\right)}{dx}}\\{{q}_{\mathrm{1}} \left({x}\right)=\mathrm{gcd}\left({q}\left({x}\right),{q}'\left({x}\right)\right)}\\{{q}_{\mathrm{2}} \left({x}\right)=\frac{{q}\left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}}\\{{p}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{3};\:{q}\left({x}\right)=\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\\{{q}'\left({x}\right)=\mathrm{8}{x}^{\mathrm{3}} \left({x}^{\mathrm{4}} +\mathrm{1}\right);\:{q}_{\mathrm{1}} \left({x}\right)={q}_{\mathrm{2}} \left({x}\right)={x}^{\mathrm{4}} +\mathrm{1}}\\{\mathrm{now}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}}\\{{p}_{\mathrm{1}} \left({x}\right)\:\mathrm{and}\:{p}_{\mathrm{2}} \left({x}\right)\:\mathrm{with}\:\mathrm{degree}\left({p}_{{n}} \right)<\mathrm{degree}\left({q}_{{n}} \right)}\\{\mathrm{using}\:\mathrm{this}\:\mathrm{equation}\:\left(\mathrm{comparing}\:\mathrm{factors}\right)}\\{\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\frac{{d}}{{dx}}\left[\frac{{p}_{\mathrm{1}} \left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}\right]+\frac{{p}_{\mathrm{2}} \left({x}\right)}{{q}_{\mathrm{2}} \left({x}\right)}}\\{{p}_{\mathrm{1}} \left({x}\right)={a}_{\mathrm{1}} {x}^{\mathrm{3}} +{b}_{\mathrm{1}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{d}_{\mathrm{1}} }\\{{p}_{\mathrm{2}} \left({x}\right)={a}_{\mathrm{2}} {x}^{\mathrm{3}} +{b}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} }\\{\Rightarrow\:{p}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right);\:{p}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({x}^{\mathrm{2}} +\mathrm{9}\right)}\end{bmatrix} \\ $$$$=\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}\left({x}^{\mathrm{4}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}^{\mathrm{2}} +\mathrm{9}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{x}^{\mathrm{2}} +\mathrm{9}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}=\int\frac{{x}^{\mathrm{2}} +\mathrm{9}}{\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\left(\frac{\mathrm{8}{x}+\mathrm{9}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}−\frac{\mathrm{8}{x}−\mathrm{9}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\right){dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\left(\mathrm{4}\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\mathrm{5}\sqrt{\mathrm{2}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}−\mathrm{4}\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\mathrm{5}\sqrt{\mathrm{2}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\right){dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{terms}\:\mathrm{are}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}}\\{{F}\left({x}\right)=\int{a}\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}{dx}\:\Rightarrow\:{F}\left({x}\right)={a}\mathrm{ln}\left({f}\left({x}\right)\right)}\\{\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{and}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{terms}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}}\\{{u}=\sqrt{\mathrm{2}}{x}\pm\mathrm{1}\:\rightarrow\:{dx}=\frac{{du}}{\sqrt{\mathrm{2}}}\:\mathrm{and}\:\mathrm{lead}\:\mathrm{to}}\\{{F}\left({x}\right)=\sqrt{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}}\:\Rightarrow\:{F}\left({x}\right)=\sqrt{\mathrm{2}}\mathrm{arctan}\left({u}\right)}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}}\left(\mathrm{ln}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)−\mathrm{ln}\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{arctan}\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)+\mathrm{arctan}\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$=\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}\left({x}^{\mathrm{4}} +\mathrm{1}\right)}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\left(\frac{\mid{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\mid}{\mid{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\mid}\right)+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}\left(\mathrm{arctan}\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)+\mathrm{arctan}\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\right)+{C} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}\pi \\ $$

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