fnd-dx-x-2-3-x-2- – Tinku Tara

Question Number 65920 by mathmax by abdo last updated on 05/Aug/19

$${fnd}\:\int\:\frac{{dx}}{{x}+\mathrm{2}−\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }} \\ $$

Commented by mathmax by abdo last updated on 12/Aug/19

let I =∫ (dx/(x+2−(√(3+x^2 )))) cha7gement x=(√3)sh(t)give I =∫ (((√3)ch(t))/( (√3)sh(t)+2−(√3)ch(t)))dt =∫ (((√3)((e^t +e^(−t) )/2))/( (√3)((e^t −e^(−t) )/2)+2−(√3)((e^t +e^(−t) )/2)))dt =∫ (((√3)(e^t +e^(−t) ))/( (√3)(e^t −e^(−t) )+4−(√3)(e^t +e^(−t) )))dt =_(e^t =u) ∫ (((√3)(u+u^(−1) ))/( (√3)(u−u^(−1) )+4−(√3)(u+u^(−1) )))(du/u) =∫ (((√3)(u+u^(−1) ))/( (√3)u^2 −(√3) +4u −(√3)u^2 −(√3))) du =∫ (((√3)(u+u^(−1) ))/(4u−2(√3)))du =∫ (((√3)u^2 +(√3))/(4u^2 −2(√3)u))du =((√3)/4) ∫ ((u^2 +1)/(u^2 −((√3)/2)))du =((√3)/4)∫ ((u^2 −((√3)/2)+1−((√3)/2))/(u^2 −((√3)/2)))du =((√3)/4)u +((√3)/8)(2−(√3)) ∫ (du/(u^2 −((√3)/2))) =((√3)/4)u +((√3)/8)(2−(√3)) ∫ ((1/(u−(√((√3)/2)))) −(1/(u+(√((√3)/2)))))du =((√3)/4)u +((√3)/8)(2−(√3))ln∣((u−(√((√3)/2)))/(u+(√((√3)/2))))∣ +c we have u=e^t and t=argsh((x/( (√3))))=ln((x/( (√3))) +(√(1+(x^2 /3)))) ⇒u=((x+(√(3+x^2 )))/( (√3))) ⇒I =(((√3)(x+(√(3+x^2 ))))/(4(√3))) +((√3)/8)(2−(√3))ln∣((((x+(√(3+x^2 )))/( (√3)))−(√((√3)/2)))/(((x+(√(3+x^2 )))/( (√3))) +(√((√3)/2))))∣ +C

$${let}\:{I}\:=\int\:\:\frac{{dx}}{{x}+\mathrm{2}−\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}\:\:{cha}\mathrm{7}{gement}\:\:{x}=\sqrt{\mathrm{3}}{sh}\left({t}\right){give} \\ $$$${I}\:=\int\:\:\:\:\frac{\sqrt{\mathrm{3}}{ch}\left({t}\right)}{\:\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{2}−\sqrt{\mathrm{3}}{ch}\left({t}\right)}{dt}\:=\int\:\:\frac{\sqrt{\mathrm{3}}\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}{\:\sqrt{\mathrm{3}}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}+\mathrm{2}−\sqrt{\mathrm{3}}\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}{dt} \\ $$$$=\int\:\:\:\frac{\sqrt{\mathrm{3}}\left({e}^{{t}} \:+{e}^{−{t}} \right)}{\:\sqrt{\mathrm{3}}\left({e}^{{t}} −{e}^{−{t}} \right)+\mathrm{4}−\sqrt{\mathrm{3}}\left({e}^{{t}} \:+{e}^{−{t}} \right)}{dt} \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\int\:\:\frac{\sqrt{\mathrm{3}}\left({u}+{u}^{−\mathrm{1}} \right)}{\:\sqrt{\mathrm{3}}\left({u}−{u}^{−\mathrm{1}} \right)+\mathrm{4}−\sqrt{\mathrm{3}}\left({u}+{u}^{−\mathrm{1}} \right)}\frac{{du}}{{u}} \\ $$$$=\int\:\frac{\sqrt{\mathrm{3}}\left({u}+{u}^{−\mathrm{1}} \right)}{\:\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:+\mathrm{4}{u}\:−\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\sqrt{\mathrm{3}}}\:{du}\:=\int\:\:\frac{\sqrt{\mathrm{3}}\left({u}+{u}^{−\mathrm{1}} \right)}{\mathrm{4}{u}−\mathrm{2}\sqrt{\mathrm{3}}}{du} \\ $$$$=\int\:\:\frac{\sqrt{\mathrm{3}}{u}^{\mathrm{2}} \:+\sqrt{\mathrm{3}}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{u}}{du}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\int\:\:\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{du} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\int\:\:\frac{{u}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{{u}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{du}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{u}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{u}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\int\:\:\:\left(\frac{\mathrm{1}}{{u}−\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\:−\frac{\mathrm{1}}{{u}+\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\right){du} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{u}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){ln}\mid\frac{{u}−\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}{{u}+\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\mid\:+{c}\:{we}\:{have} \\ $$$${u}={e}^{{t}} \:\:{and}\:{t}={argsh}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)={ln}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:\Rightarrow{u}=\frac{{x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{I}\:=\frac{\sqrt{\mathrm{3}}\left({x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }\right)}{\mathrm{4}\sqrt{\mathrm{3}}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){ln}\mid\frac{\frac{{x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}−\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}{\frac{{x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\mid\:+{C} \\ $$

Answered by MJS last updated on 07/Aug/19

∫(dx/(x+2−(√(x^2 +3))))=2∫(dx/(4x+1))+∫(x/(4x+1))dx+3∫((√(x^2 +3))/(4x+1))dx 2∫(dx/(4x+1))=(1/2)ln (4x+1) ∫(x/(4x+1))dx=(1/4)∫dx−(1/4)∫(dx/(4x+1))=(x/4)−(1/(16))ln (4x+1) 3∫((√(x^2 +3))/(4x+1))dx= [t=arctan (((√3)/3)x) → dx=((√3)/3)(x^2 +3)dt] =9∫(dt/((4(√3)sin t +cos t)cos^2 t))= [u=tan (t/2) → dt=2cos^2 (t/2) du] =−18∫(((u^2 +1)^2 )/((u−1)^2 (u+1)^2 (u−7−4(√3))(u+7−4(√3))))du= =((3(1−4(√3)))/(16))∫(du/(u−1))+((3(√3))/4)∫(u/((u−1)^2 ))du−((3(1+4(√3)))/(16))∫(du/(u+1))+((3(√3))/4)∫(u/((u+1)^2 ))du−((21)/(16))∫(du/(u−7−4(√3)))+((21)/(16))∫(du/(u+7−4(√3)))= now it′s easy... =((21)/(16))ln ((u+7−4(√3))/(u−7−4(√3))) +(3/(16))ln ((u−1)/(u+1)) −((3(√3))/(2u^2 −2))= [u=tan (t/2) =(((√(x^2 +3))−(√3))/x)] =((21)/(16))ln ∣((x−12+7(√(x^2 +3)))/(4x+1))∣ +(3/(16))ln ∣x−(√(x^2 +3))∣ +(3/4)(√(x^2 +3)) ⇒ ∫(dx/(x+2−(√(x^2 +3))))=(7/(16))ln ∣4x+1∣ +((21)/(16))ln ∣((x−12+7(√(x^2 +3)))/(4x+1))∣ +(3/(16))ln ∣x−(√(x^2 +3))∣ +(1/4)(x+3(√(x^2 +3))) +C ...I hope I made no typos...

$$\int\frac{{dx}}{{x}+\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}=\mathrm{2}\int\frac{{dx}}{\mathrm{4}{x}+\mathrm{1}}+\int\frac{{x}}{\mathrm{4}{x}+\mathrm{1}}{dx}+\mathrm{3}\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{4}{x}+\mathrm{1}}{dx} \\ $$$$\mathrm{2}\int\frac{{dx}}{\mathrm{4}{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{4}{x}+\mathrm{1}\right) \\ $$$$\int\frac{{x}}{\mathrm{4}{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\mathrm{4}{x}+\mathrm{1}}=\frac{{x}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left(\mathrm{4}{x}+\mathrm{1}\right) \\ $$$$\mathrm{3}\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{4}{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{x}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\mathrm{3}\right){dt}\right] \\ $$$$=\mathrm{9}\int\frac{{dt}}{\left(\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}\:{t}\:+\mathrm{cos}\:{t}\right)\mathrm{cos}^{\mathrm{2}} \:{t}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\mathrm{2cos}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}\:{du}\right] \\ $$$$=−\mathrm{18}\int\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left({u}−\mathrm{1}\right)^{\mathrm{2}} \left({u}+\mathrm{1}\right)^{\mathrm{2}} \left({u}−\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\left({u}+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)}{du}= \\ $$$$=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{4}\sqrt{\mathrm{3}}\right)}{\mathrm{16}}\int\frac{{du}}{{u}−\mathrm{1}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\int\frac{{u}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }{du}−\frac{\mathrm{3}\left(\mathrm{1}+\mathrm{4}\sqrt{\mathrm{3}}\right)}{\mathrm{16}}\int\frac{{du}}{{u}+\mathrm{1}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\int\frac{{u}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{du}−\frac{\mathrm{21}}{\mathrm{16}}\int\frac{{du}}{{u}−\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}+\frac{\mathrm{21}}{\mathrm{16}}\int\frac{{du}}{{u}+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}= \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}… \\ $$$$=\frac{\mathrm{21}}{\mathrm{16}}\mathrm{ln}\:\frac{{u}+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}{{u}−\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\:−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{\mathrm{3}}}{{x}}\right] \\ $$$$=\frac{\mathrm{21}}{\mathrm{16}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{12}+\mathrm{7}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{4}{x}+\mathrm{1}}\mid\:+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{ln}\:\mid{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\mid\:+\frac{\mathrm{3}}{\mathrm{4}}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\int\frac{{dx}}{{x}+\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}=\frac{\mathrm{7}}{\mathrm{16}}\mathrm{ln}\:\mid\mathrm{4}{x}+\mathrm{1}\mid\:+\frac{\mathrm{21}}{\mathrm{16}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{12}+\mathrm{7}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{4}{x}+\mathrm{1}}\mid\:+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{ln}\:\mid{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\mid\:+\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{3}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)\:+{C} \\ $$$$…\mathrm{I}\:\mathrm{hope}\:\mathrm{I}\:\mathrm{made}\:\mathrm{no}\:\mathrm{typos}… \\ $$

Commented by mathmax by abdo last updated on 07/Aug/19

thank you sir mjs for this hard word but you can use the changement x =(√3)sh(t)...its eazy in this case...

$${thank}\:{you}\:{sir}\:{mjs}\:{for}\:{this}\:{hard}\:{word}\:\:{but}\:{you}\:{can}\:{use}\:{the} \\ $$$${changement}\:{x}\:=\sqrt{\mathrm{3}}{sh}\left({t}\right)…{its}\:{eazy}\:{in}\:{this}\:{case}… \\ $$

Commented by MJS last updated on 07/Aug/19

true... but it was great fun to solve it like this...

$$\mathrm{true}… \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{was}\:\mathrm{great}\:\mathrm{fun}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{like}\:\mathrm{this}… \\ $$

Commented by MJS last updated on 07/Aug/19

I then get (7/8)ln ∣2x−3+2(√(x^2 +3))∣ −(1/2)arcsinh (((√3)/3)x) +(1/4)(x+(√(x^2 +3)))+C

$$\mathrm{I}\:\mathrm{then}\:\mathrm{get} \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}\mathrm{ln}\:\mid\mathrm{2}{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsinh}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{x}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)+{C} \\ $$

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