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Question Number 37018 by behi83417@gmail.com last updated on 08/Jun/18
Commented by math khazana by abdo last updated on 08/Jun/18
1) let I = ∫ ((√(1+x^2 ))/(1+x))dx changement x=sht give I = ∫ ((ch(t))/(1+sh(t)))ch(t)dt= ∫ ((ch^2 (t)dt)/(1+sh(t))) =(1/2) ∫ ((1+ch(2t))/(1+sh(t)))dt =(1/2) ∫ ((1 +((e^(2t) +e^(−2t) )/2))/(1+((e^t −e^(−t) )/2)))dt 2I = ∫ ((2 +e^(2t) +e^(−2t) )/(2 +e^t −e^(−t) ))dt =_(e^t =u) ∫ ((2 +u^2 +(1/u^2 ))/(2 +u −(1/u))) du =∫ ((2u^2 +u^4 +1)/(2u +u^2 −1))(1/u)du =∫ ((u^4 +2u^2 +1)/(u^3 +2u^2 −u))du = ∫ ((u(u^3 +2u^2 −u) −2u^3 +u^2 +2u^2 +1)/(u^3 +2u^2 −u)) du = (u^2 /2) +∫ ((−2u^3 +3u^2 +1)/(u^3 +2u^2 −u))du =(u^2 /2) +∫ ((−2( u^3 +2u^2 −u) +4u^2 −2u +3u^2 +1)/(u^3 +2u^2 −u))du =(u^2 /2) −2u +∫ ((7u^2 −2u +1)/(u(u^2 +2u−1)))du let?decompose F(u)= ((7u^2 −2u +1)/(u(u^2 +2u−1))) roots of u^2 +2u−1 Δ^′ = 2 ⇒ u_1 =−1+(√2) and u_2 =−1−(√2) F(u) = ((7u^2 −2u +1)/(u(u−u_1 )(u−u_2 ))) = (a/u) +(b/(u−u_1 )) +(c/(u−u_2 )) its simple to find a ,b and c so ∫ F(u)du =aln∣u∣ +bln∣u−u_1 ∣ +cln∣u−u_2 ∣ +C =at +b ln∣ e^t −u_1 ∣ +c ln∣ e^t −u_2 ∣ +C =aln(x+(√(1+x^2 )) ) +bln∣ x+(√(1+x^2 ))−u_1 ∣ +c ln∣ x+(√(1+x^2 )) −u_2 ∣ +C = 2I ⇒ I =(a/2)ln(x+(√(1+x^2 ))) +(b/2)ln∣x+(√(1+x^2 )) −u_1 ∣ +(c/2)ln∣x+(√(1+x^2 )) −u_2 ∣ +C .
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\:\int\:\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+{x}}{dx}\:{changement}\:{x}={sht}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\frac{{ch}\left({t}\right)}{\mathrm{1}+{sh}\left({t}\right)}{ch}\left({t}\right){dt}=\:\int\:\:\:\frac{{ch}^{\mathrm{2}} \left({t}\right){dt}}{\mathrm{1}+{sh}\left({t}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{sh}\left({t}\right)}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{1}\:+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}}{\mathrm{1}+\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}}{dt} \\ $$$$\mathrm{2}{I}\:\:=\:\:\int\:\:\:\frac{\mathrm{2}\:+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}\:+{e}^{{t}} \:−{e}^{−{t}} }{dt}\: \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\int\:\:\:\:\frac{\mathrm{2}\:\:+{u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{\mathrm{2}\:+{u}\:−\frac{\mathrm{1}}{{u}}}\:{du}\:=\int\:\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+{u}^{\mathrm{4}} \:+\mathrm{1}}{\mathrm{2}{u}\:+{u}^{\mathrm{2}} \:−\mathrm{1}}\frac{\mathrm{1}}{{u}}{du} \\ $$$$=\int\:\:\:\:\frac{{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} \:−{u}}{du} \\ $$$$=\:\int\:\:\:\frac{{u}\left({u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} −{u}\right)\:−\mathrm{2}{u}^{\mathrm{3}} \:+{u}^{\mathrm{2}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} \:−{u}}\:{du} \\ $$$$=\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\int\:\:\:\frac{−\mathrm{2}{u}^{\mathrm{3}} \:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} \:−{u}}{du} \\ $$$$=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\int\:\:\:\frac{−\mathrm{2}\left(\:{u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} \:−{u}\right)\:+\mathrm{4}{u}^{\mathrm{2}} \:−\mathrm{2}{u}\:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{3}} \:+\mathrm{2}{u}^{\mathrm{2}} \:−{u}}{du} \\ $$$$=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:−\mathrm{2}{u}\:\:+\int\:\:\:\:\frac{\mathrm{7}{u}^{\mathrm{2}} \:−\mathrm{2}{u}\:+\mathrm{1}}{{u}\left({u}^{\mathrm{2}} \:+\mathrm{2}{u}−\mathrm{1}\right)}{du}\:{let}?{decompose} \\ $$$${F}\left({u}\right)=\:\frac{\mathrm{7}{u}^{\mathrm{2}} −\mathrm{2}{u}\:+\mathrm{1}}{{u}\left({u}^{\mathrm{2}} \:+\mathrm{2}{u}−\mathrm{1}\right)}\:{roots}\:{of}\:{u}^{\mathrm{2}} \:+\mathrm{2}{u}−\mathrm{1} \\ $$$$\Delta^{'} \:=\:\mathrm{2}\:\Rightarrow\:{u}_{\mathrm{1}} =−\mathrm{1}+\sqrt{\mathrm{2}}\:\:{and}\:{u}_{\mathrm{2}} =−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$${F}\left({u}\right)\:=\:\frac{\mathrm{7}{u}^{\mathrm{2}} \:−\mathrm{2}{u}\:+\mathrm{1}}{{u}\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)}\:=\:\frac{{a}}{{u}}\:\:+\frac{{b}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{c}}{{u}−{u}_{\mathrm{2}} } \\ $$$${its}\:{simple}\:{to}\:{find}\:{a}\:,{b}\:{and}\:{c}\:{so} \\ $$$$\int\:{F}\left({u}\right){du}\:={aln}\mid{u}\mid\:+{bln}\mid{u}−{u}_{\mathrm{1}} \mid\:+{cln}\mid{u}−{u}_{\mathrm{2}} \mid\:+{C} \\ $$$$={at}\:+{b}\:{ln}\mid\:{e}^{{t}} \:−{u}_{\mathrm{1}} \mid\:+{c}\:{ln}\mid\:{e}^{{t}} \:−{u}_{\mathrm{2}} \mid\:+{C} \\ $$$$={aln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)\:+{bln}\mid\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{u}_{\mathrm{1}} \mid \\ $$$$+{c}\:{ln}\mid\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−{u}_{\mathrm{2}} \mid\:+{C}\:=\:\mathrm{2}{I}\:\Rightarrow \\ $$$${I}\:=\frac{{a}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{b}}{\mathrm{2}}{ln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−{u}_{\mathrm{1}} \mid \\ $$$$+\frac{{c}}{\mathrm{2}}{ln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−{u}_{\mathrm{2}} \mid\:+{C}\:. \\ $$
Commented by behi83417@gmail.com last updated on 08/Jun/18
∫ ((√(1+x^2 ))/(1+x))dx= (√2)ln((((√(1+x^2 ))−x−(√2)−1)/( (√(1+x^2 ))−x+(√2)−1)))+ln((√(1+x^2 ))−x)+(√(1+x^2 ))+C
$$\int\:\frac{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{1}+\boldsymbol{{x}}}\boldsymbol{{dx}}= \\ $$$$\sqrt{\mathrm{2}}\boldsymbol{{ln}}\left(\frac{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }−\boldsymbol{{x}}−\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }−\boldsymbol{{x}}+\sqrt{\mathrm{2}}−\mathrm{1}}\right)+\boldsymbol{{ln}}\left(\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }−\boldsymbol{{x}}\right)+\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }+\boldsymbol{{C}} \\ $$
Commented by MJS last updated on 08/Jun/18
sorry but this is wrong ∫ ((√(1+x^2 ))/(1+x))dx= [t=arctan x → dx=sec^2 t dt] =∫((sec^2 t (√(1+tan^2 t)))/(1+tan t))dt=∫((sec^3 t)/(1+tan t))= [Weierstrass: u=tan (t/2) → dt=((2du)/(1+u^2 ))] =−2∫(((u^2 +1)^2 )/((u−1)^2 (u+1)^2 (u^2 −2u−1)))du= =−4∫(du/(u^2 −2u−1))+ +∫(du/((u−1)^2 ))−∫(du/((u+1)^2 ))+ +∫(du/(u−1))−∫(du/(u+1))= =−(√2)∫(du/(u−(√2)−1))+(√2)∫(du/(u+(√2)−1))− −(1/(u−1))+(1/(u+1))+ +ln(u−1)−ln(u+1)= =−(√2)ln(u−(√2)−1)+(√2)ln(u+(√2)−1)+ +(2/(1−u^2 ))+ln ((u−1)/(u+1))= =(√2)ln ((u+(√2)−1)/(u−(√2)−1))+ln ((u−1)/(u+1))+(2/(1−u^2 ))= =(√2)ln ((tan (t/2) +(√2)−1)/(tan (t/2) −(√2)−1))+ln ((tan (t/2) −1)/(tan (t/2) +1))+(2/(1−tan^2 (t/2)))= [tan ((arctan x)/2)=(((√(x^2 +1))−1)/x)] =(√2)ln (((√(x^2 +1))−(1−(√2))x−1)/( (√(x^2 +1))−(1+(√2))x−1))+ln (((√(x^2 +1))−x−1)/( (√(x^2 +1))+x−1))+(x^2 /( (√(x^2 +1))−1))= =(√2)ln (((√(x^2 +1))−(1−(√2))x−1)/( (√(x^2 +1))−(1+(√2))x−1))+ln(x−(√(x^2 +1)))+(√(x^2 +1))+1= =(√2)ln (((1−(√2))(x−1)−(2−(√2))(√(x^2 +1)))/(x+1))+ln(x−(√(x^2 +1)))+(√(x^2 +1))+C
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\int\:\frac{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{1}+\boldsymbol{{x}}}\boldsymbol{{dx}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{arctan}\:{x}\:\rightarrow\:{dx}=\mathrm{sec}^{\mathrm{2}} \:{t}\:{dt}\right] \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{2}} \:{t}\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{t}}}{\mathrm{1}+\mathrm{tan}\:{t}}{dt}=\int\frac{\mathrm{sec}^{\mathrm{3}} \:{t}}{\mathrm{1}+\mathrm{tan}\:{t}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{Weierstrass}:\:{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\right] \\ $$$$=−\mathrm{2}\int\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left({u}−\mathrm{1}\right)^{\mathrm{2}} \left({u}+\mathrm{1}\right)^{\mathrm{2}} \left({u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{1}\right)}{du}= \\ $$$$=−\mathrm{4}\int\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{1}}+ \\ $$$$\:\:\:\:\:+\int\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }−\int\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\int\frac{{du}}{{u}−\mathrm{1}}−\int\frac{{du}}{{u}+\mathrm{1}}= \\ $$$$=−\sqrt{\mathrm{2}}\int\frac{{du}}{{u}−\sqrt{\mathrm{2}}−\mathrm{1}}+\sqrt{\mathrm{2}}\int\frac{{du}}{{u}+\sqrt{\mathrm{2}}−\mathrm{1}}− \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{{u}−\mathrm{1}}+\frac{\mathrm{1}}{{u}+\mathrm{1}}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{ln}\left({u}−\mathrm{1}\right)−\mathrm{ln}\left({u}+\mathrm{1}\right)= \\ $$$$=−\sqrt{\mathrm{2}}\mathrm{ln}\left({u}−\sqrt{\mathrm{2}}−\mathrm{1}\right)+\sqrt{\mathrm{2}}\mathrm{ln}\left({u}+\sqrt{\mathrm{2}}−\mathrm{1}\right)+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{2}}{\mathrm{1}−{u}^{\mathrm{2}} }+\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{{u}+\sqrt{\mathrm{2}}−\mathrm{1}}{{u}−\sqrt{\mathrm{2}}−\mathrm{1}}+\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{1}−{u}^{\mathrm{2}} }= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:−\sqrt{\mathrm{2}}−\mathrm{1}}+\mathrm{ln}\:\frac{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:−\mathrm{1}}{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{tan}\:\frac{\mathrm{arctan}\:{x}}{\mathrm{2}}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}}{{x}}\right] \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}−\mathrm{1}}+\mathrm{ln}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}−\mathrm{1}}+\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}}= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}−\mathrm{1}}+\mathrm{ln}\left({x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({x}−\mathrm{1}\right)−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}+\mathrm{ln}\left({x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{C} \\ $$
Answered by MJS last updated on 08/Jun/18
2) ∫((√x)/(x^3 +x^2 +x+1))dx= [t=(√x) → dx=2(√x)dt] =2∫(t^2 /(t^6 +t^4 +t^2 +1))dt=2∫(t^2 /((t^4 +1)(t^2 +1)))dt= =∫((t^2 +1)/(t^4 +1))dt−∫(dt/(t^2 +1))=∫((t^2 +1)/(t^4 +1))dt−arctan t= ∫((t^2 +1)/(t^4 +1))dt=∫((t^2 +1)/((t^2 −(√2)t+1)(t^2 +(√2)t+1)))dt= =(1/2)∫(dt/(t^2 −(√2)t+1))+(1/2)∫(dt/(t^2 +(√2)t+1))= =(1/2)∫(dt/((t−((√2)/2))^2 +(1/2)))+(1/2)∫(dt/((t+((√2)/2))^2 +(1/2)))= [u=(√2)t±1 → dt=((√2)/2)du] =((√2)/2)∫(du_− /(u_− ^2 +1))+((√2)/2)∫(du_+ /(u_+ ^2 +1))= =((√2)/2)arctan u_− +((√2)/2)arctan u_+ = =((√2)/2)arctan((√2)t−1)+((√2)/2)arctan((√2)t+1) =((√2)/2)arctan((√2)t−1)+((√2)/2)arctan((√2)t+1)−arctan t= =((√2)/2)(arctan((√(2x))−1)+arctan((√(2x))+1))−arctan (√x)+C
$$\left.\mathrm{2}\right) \\ $$$$\int\frac{\sqrt{{x}}}{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} +{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{4}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}−\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}−\mathrm{arctan}\:{t}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left({t}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left({t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}=\sqrt{\mathrm{2}}{t}\pm\mathrm{1}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{du}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{du}_{−} }{{u}_{−} ^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{du}_{+} }{{u}_{+} ^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:{u}_{−} \:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:{u}_{+} = \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$$ \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)−\mathrm{arctan}\:{t}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{arctan}\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right)+\mathrm{arctan}\left(\sqrt{\mathrm{2}{x}}+\mathrm{1}\right)\right)−\mathrm{arctan}\:\sqrt{{x}}+{C} \\ $$
Answered by MJS last updated on 08/Jun/18
1) I messed around a little, found this: ∫((√(1+x^2 ))/(1+x))dx= =(√2)arcsinh(−((1−x)/(∣1+x∣)))−arcsinh x +(√(1+x^2 ))+C let me explain: ∫((√(1+x^2 ))/(1+x))dx=∫((1+x^2 )/((1+x)(√(1+x^2 ))))dx=∫((2+x^2 −1)/((1+x)(√(1+x^2 ))))dx= =2∫(dx/((1+x)(√(1+x^2 ))))−∫((1−x^2 )/((1+x)(√(1+x^2 ))))dx= =2∫(dx/((1+x)(√(1+x^2 ))))−∫(((1−x)(1+x))/((1+x)(√(1+x^2 ))))dx= =2∫(dx/((1+x)(√(1+x^2 ))))−∫((1−x)/( (√(1+x^2 ))))dx= =2∫(dx/((1+x)(√(1+x^2 ))))−∫(dx/( (√(1+x^2 ))))+∫(x/( (√(1+x^2 ))))dx= [the 2^(nd) and 3^(rd) ones are standard integrals] =2∫(dx/((1+x)(√(1+x^2 ))))−arcsinh x +(√(1+x^2 )) now consider this: (d/dx)[arcsinh x]=(1/( (√(1+x^2 )))) (d/dx)[arcsinh f(x)]=((f′(x))/( (√(1+f^2 (x))))) let f(x)=((g(x))/(h(x)))=(u/v) ⇒ ⇒ ((f′(x))/( (√(1+f^2 (x)))))=(((u′v−uv′)/v^2 )/( (√(1+(u/v)^2 ))))=((u′v−uv′)/(∣v∣(√(u^2 +v^2 )))) let v=1+x ⇒ ⇒ ((u′v−uv′)/(∣v∣(√(u^2 +v^2 ))))=((u′(1+x)−u)/(∣1+x∣(√(u^2 +1+2x+x^2 )))) we′d like the following equations to be true: (A) u′(1+x)−u=2 (B) (√(u^2 +1+2x+x^2 ))=(√(1+x^2 )) (A) works with u=x−1 but (B) (√(u^2 +1+2x+x^2 ))= =(√((x−1)^2 +1+2x+x^2 ))=(√(2(1+x^2 ))) we now have: (d/dx)[arcsinh(−((1−x)/(1+x)))]=((√2)/(∣1+x∣(√(1+x^2 )))) ⇒ ⇒ 2∫(dx/(∣1+x∣(√(1+x^2 ))))=(√2)arcsinh(−((1−x)/(1+x))) ⇒ [F(x)=arcsin(−((1−x)/(1+x))) has an uneven pole at x=−1, F′(x) is increasing for −∞<x<∞ F^� (x)=arcsin(−((1−x)/(∣1+x∣))) has an even pole at x=−1, F^� ′(x) is increasing for −∞<x<−1 and decreasing for −1<x<∞] ⇒ 2∫(dx/((1+x)(√(1+x^2 ))))=(√2)arcsinh(−((1−x)/(∣1+x∣)))
$$\left.\mathrm{1}\right)\:\mathrm{I}\:\mathrm{messed}\:\mathrm{around}\:\mathrm{a}\:\mathrm{little},\:\mathrm{found}\:\mathrm{this}: \\ $$$$ \\ $$$$\int\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+{x}}{dx}= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{arcsinh}\left(−\frac{\mathrm{1}−{x}}{\mid\mathrm{1}+{x}\mid}\right)−\mathrm{arcsinh}\:{x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{C} \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{me}\:\mathrm{explain}: \\ $$$$ \\ $$$$\int\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+{x}}{dx}=\int\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=\int\frac{\mathrm{2}+{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}= \\ $$$$=\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\int\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}= \\ $$$$=\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\int\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}= \\ $$$$=\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\int\frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}= \\ $$$$=\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}+\int\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{and}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{ones}\:\mathrm{are}\:\mathrm{standard}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\mathrm{integrals}\right] \\ $$$$=\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\mathrm{arcsinh}\:{x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{consider}\:\mathrm{this}: \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{arcsinh}\:{x}\right]=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{arcsinh}\:{f}\left({x}\right)\right]=\frac{{f}'\left({x}\right)}{\:\sqrt{\mathrm{1}+{f}^{\mathrm{2}} \left({x}\right)}} \\ $$$$\mathrm{let}\:{f}\left({x}\right)=\frac{{g}\left({x}\right)}{{h}\left({x}\right)}=\frac{{u}}{{v}}\:\Rightarrow \\ $$$$\Rightarrow\:\frac{{f}'\left({x}\right)}{\:\sqrt{\mathrm{1}+{f}^{\mathrm{2}} \left({x}\right)}}=\frac{\frac{{u}'{v}−{uv}'}{{v}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\left({u}/{v}\right)^{\mathrm{2}} }}=\frac{{u}'{v}−{uv}'}{\mid{v}\mid\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }} \\ $$$$\mathrm{let}\:{v}=\mathrm{1}+{x}\:\Rightarrow \\ $$$$\Rightarrow\:\frac{{u}'{v}−{uv}'}{\mid{v}\mid\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }}=\frac{{u}'\left(\mathrm{1}+{x}\right)−{u}}{\mid\mathrm{1}+{x}\mid\sqrt{{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }} \\ $$$$\mathrm{we}'\mathrm{d}\:\mathrm{like}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equations}\:\mathrm{to}\:\mathrm{be}\:\mathrm{true}: \\ $$$$\left({A}\right)\:\:\:\:\:{u}'\left(\mathrm{1}+{x}\right)−{u}=\mathrm{2} \\ $$$$\left({B}\right)\:\:\:\:\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left({A}\right)\:\mathrm{works}\:\mathrm{with}\:{u}={x}−\mathrm{1}\:\mathrm{but}\:\left({B}\right)\:\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }= \\ $$$$=\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\mathrm{we}\:\mathrm{now}\:\mathrm{have}: \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{arcsinh}\left(−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\right]=\frac{\sqrt{\mathrm{2}}}{\mid\mathrm{1}+{x}\mid\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{2}\int\frac{{dx}}{\mid\mathrm{1}+{x}\mid\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\sqrt{\mathrm{2}}\mathrm{arcsinh}\left(−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{F}\left({x}\right)=\mathrm{arcsin}\left(−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:\mathrm{has}\:\mathrm{an}\:\mathrm{uneven}\:\mathrm{pole}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{at}\:{x}=−\mathrm{1},\:{F}'\left({x}\right)\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{for}\:−\infty<{x}<\infty \\ $$$$\:\:\:\:\:\:\:\:\:\:\bar {{F}}\left({x}\right)=\mathrm{arcsin}\left(−\frac{\mathrm{1}−{x}}{\mid\mathrm{1}+{x}\mid}\right)\:\mathrm{has}\:\mathrm{an}\:\mathrm{even}\:\mathrm{pole} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{at}\:{x}=−\mathrm{1},\:\bar {{F}}'\left({x}\right)\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{for}\:−\infty<{x}<−\mathrm{1} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\mathrm{decreasing}\:\mathrm{for}\:−\mathrm{1}<{x}<\infty\right] \\ $$$$\Rightarrow\:\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\sqrt{\mathrm{2}}\mathrm{arcsinh}\left(−\frac{\mathrm{1}−{x}}{\mid\mathrm{1}+{x}\mid}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18
2)∫_0 ^1 (((√x) )/(1+x+x^2 +x^3 ))dx t^2 =x dx=2tdt ∫_0 ^1 ((t×2tdt)/(1+t^2 +t^4 +t^6 )) =2∫_0 ^1 (t^2 /((1+t^2 )+t^4 (1+t^2 )))dt =2∫_0 ^1 (t^2 /((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (((t^2 +t^4 )−(1+t^4 )+1)/((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (t^2 /(1+t^4 ))dt−2∫_0 ^1 (dt/(1+t^2 ))+2∫_0 ^1 (1/((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (dt/(t^2 +(1/t^2 )))−2∫_0 ^1 (dt/(1+t^2 ))+2∫_0 ^1 (dt/(t(t+(1/t))t^2 (t^2 +(1/t^2 )))) I_1 =∫_0 ^1 ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =∫_0 ^1 ((d(t+(1/t)))/((t+(1/t))^2 −2))+∫_0 ^1 ((d(t−(1/t)))/((t−(1/t))^2 +2)) now uzeformula ∫(dk/(k^2 −a^2 )) and ∫(dp/(p^2 +a^2 )) ={(1/(2(√2) ))ln∣((t+(1/t)−(√2) )/(t+(1/t)+(√2) ))∣+(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2))))}_0 ^1 =I_1 I_2 =−2∫_0 ^1 (dt/(1+t^2 )) ={−2×tan^(−1) (t)}_0 ^1 =I_2
$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{{x}}\:}{\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{dx} \\ $$$${t}^{\mathrm{2}} ={x}\:\:\:{dx}=\mathrm{2}{tdt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}×\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{2}} +{t}^{\mathrm{4}} +{t}^{\mathrm{6}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{t}^{\mathrm{4}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right)−\left(\mathrm{1}+{t}^{\mathrm{4}} \right)+\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}\left({t}+\frac{\mathrm{1}}{{t}}\right){t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$${now}\:{uzeformula}\:\int\frac{{dk}}{{k}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\:\:{and}\:\int\frac{{dp}}{{p}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$=\left\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\mid\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}\:}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}\:}\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} ={I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\left\{−\mathrm{2}×{tan}^{−\mathrm{1}} \left({t}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} \:\:={I}_{\mathrm{2}} \\ $$$$ \\ $$

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