Question Number 71665 by mathmax by abdo last updated on 18/Oct/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left(\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)} \\ $$
Answered by MJS last updated on 19/Oct/19
![∫(dx/((x+1)^2 (√(x^2 +4))))= [t=arctan (x/2) → dx=((x^2 +4)/2)dt] =∫((cos t)/((cos t +2sin t)^2 ))dt= [u=tan (t/2) → dt=((2du)/(u^2 +1))] =−2∫((u^2 −1)/((u^2 −4u−1)^2 )) now I′d use Ostrogradski′s Method once again ∫(P/Q)=(P_1 /Q_1 )+∫(P_2 /Q_2 ) Q_1 =gcd (Q, Q′); Q_2 =(Q/Q_1 ) degree (P_i ) <degree (Q_i ) we find P_i by comparing the factors of (P/Q)=((P_1 /Q_1 ))′+(P_2 /Q_2 ) Q=(u^2 −4u−1)^2 Q′=4(u−2)(u^2 −4u−1) Q_1 =Q_2 =u^2 −4u−1 P=u^2 −1 P_1 =−(4/5)u−(2/5) P_2 =(1/5) −2∫((u^2 −1)/((u^2 −4u−1)^2 ))= =((4(2u+1))/(5(u^2 −4u−1)))−(2/5)∫(du/(u^2 −4u−1))= =((4(2u+1))/(5(u^2 −4u−1)))−((√5)/(25))ln ((u−2−(√5))/(u−2+(√5))) = =...=−((√(x^2 +4))/(5(x+1)))+((√5)/(25))ln ∣((x−4−(√(5(x^2 +4))))/(x+1))∣ +C ∫_0 ^∞ (dx/((x+1)^2 (√(x^2 +4))))=(1/5)+((√5)/(25))ln (((7−3(√5))/2))](https://www.tinkutara.com/question/Q71700.png)
$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arctan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}}{dt}\right] \\ $$$$=\int\frac{\mathrm{cos}\:{t}}{\left(\mathrm{cos}\:{t}\:+\mathrm{2sin}\:{t}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=−\mathrm{2}\int\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{now}\:\mathrm{I}'\mathrm{d}\:\mathrm{use}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{once}\:\mathrm{again} \\ $$$$\int\frac{{P}}{{Q}}=\frac{{P}_{\mathrm{1}} }{{Q}_{\mathrm{1}} }+\int\frac{{P}_{\mathrm{2}} }{{Q}_{\mathrm{2}} } \\ $$$${Q}_{\mathrm{1}} =\mathrm{gcd}\:\left({Q},\:{Q}'\right);\:{Q}_{\mathrm{2}} =\frac{{Q}}{{Q}_{\mathrm{1}} } \\ $$$$\mathrm{degree}\:\left({P}_{{i}} \right)\:<\mathrm{degree}\:\left({Q}_{{i}} \right) \\ $$$$\mathrm{we}\:\mathrm{find}\:{P}_{{i}} \:\mathrm{by}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\frac{{P}}{{Q}}=\left(\frac{{P}_{\mathrm{1}} }{{Q}_{\mathrm{1}} }\right)'+\frac{{P}_{\mathrm{2}} }{{Q}_{\mathrm{2}} } \\ $$$${Q}=\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Q}'=\mathrm{4}\left({u}−\mathrm{2}\right)\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right) \\ $$$${Q}_{\mathrm{1}} ={Q}_{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1} \\ $$$${P}={u}^{\mathrm{2}} −\mathrm{1} \\ $$$${P}_{\mathrm{1}} =−\frac{\mathrm{4}}{\mathrm{5}}{u}−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${P}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$−\mathrm{2}\int\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{4}\left(\mathrm{2}{u}+\mathrm{1}\right)}{\mathrm{5}\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{4}\left(\mathrm{2}{u}+\mathrm{1}\right)}{\mathrm{5}\left({u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}\right)}−\frac{\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:\frac{{u}−\mathrm{2}−\sqrt{\mathrm{5}}}{{u}−\mathrm{2}+\sqrt{\mathrm{5}}}\:= \\ $$$$=…=−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{5}\left({x}+\mathrm{1}\right)}+\frac{\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{4}−\sqrt{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}}{{x}+\mathrm{1}}\mid\:+{C} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:\left(\frac{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$
Commented by Abdo msup. last updated on 19/Oct/19
$${thank}\:{you}\:{sir}. \\ $$